This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.

Alice received a set of Toy Train™ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 ≤ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.

Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ b_i).

<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.

The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.

Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 100; 1 ≤ m ≤ 200) — the number of stations and the number of candies, respectively.

The i-th of the following m lines contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — the station that initially contains candy i and the destination station of the candy, respectively.

Output

In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.

Examples

Input


5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5


Output


10 9 10 10 9 


Input


2 3
1 2
1 2
1 2


Output


5 6 

Note

Consider the second sample.

If the train started at station 1, the optimal strategy is as follows.

  1. Load the first candy onto the train. 
  2. Proceed to station 2. This step takes 1 second. 
  3. Deliver the first candy. 
  4. Proceed to station 1. This step takes 1 second. 
  5. Load the second candy onto the train. 
  6. Proceed to station 2. This step takes 1 second. 
  7. Deliver the second candy. 
  8. Proceed to station 1. This step takes 1 second. 
  9. Load the third candy onto the train. 
  10. Proceed to station 2. This step takes 1 second. 
  11. Deliver the third candy. 



Hence, the train needs 5 seconds to complete the tasks.

If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
// package Practice.CF1130;

import java.io.*;
import java.util.*;

public class CF1130D1 {
    static FastReader s;
    static PrintWriter out;
    static String INPUT = "5 7\n" +
            "2 4\n" +
            "5 1\n" +
            "2 3\n" +
            "3 4\n" +
            "4 1\n" +
            "5 3\n" +
            "3 5\n";

    public static void main(String[] args) {
        long time = System.currentTimeMillis();
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        out = new PrintWriter(System.out);
        s = new FastReader(oj);
        int n = s.nextInt();
        int m = s.nextInt();
        int[] counter = new int[n];
        int[] min = new int[n];
        for (int i = 0; i < n; ++i) {
            min[i] = Integer.MAX_VALUE;
        }
        for (int i = 0; i < m; ++i) {
            int from = s.nextInt();
            int to = s.nextInt();
            counter[from - 1]++;
            min[from - 1] = Math.min(min[from - 1], (to - from + n) % n);
        }
        for (int i = 0; i < n; ++i) {
            int max = Integer.MIN_VALUE;
            for (int j = 0; j < n; ++j) {
                if (counter[j] == 0) {
                    continue;
                }
                max = Math.max(max, (counter[j] - 1) * n + (j - i + n) % n + min[j]);
            }
            System.out.print(max + " ");
        }
        if (!oj) {
            System.out.println(Arrays.deepToString(new Object[]{System.currentTimeMillis() - time + " ms"}));
        }
        out.flush();
    }

    private static class Matrix {

        static long[][] I;
        static long mod = 1000000007;

        public static long[][] exp(long[][] M, long n) {
            if (n <= 0) {
                I = new long[M.length][M.length];
                for (int i = 0; i < M.length; i++) {
                    I[i][i] = 1L;
                }
                return I;
            }
            if (n == 1) return M;
            long[][] res = exp(M, n / 2);
            res = mult(res, res);
            if (n % 2 == 0) return res;
            return mult(res, M);
        }

        public static long[][] mult(long[][] p, long[][] q) {
            long[][] r = new long[p.length][q[0].length];
            for (int i = 0; i < p.length; i++)
                for (int j = 0; j < q[0].length; j++)
                    for (int k = 0; k < q.length; k++) {
                        r[i][j] += p[i][k] * q[k][j];
                        r[i][j] %= mod;
                    }
            return r;
        }
    }

    private static class Maths {
        static ArrayList<Long> printDivisors(long n) {
            // Note that this loop runs till square root
            ArrayList<Long> list = new ArrayList<>();
            for (long i = 1; i <= Math.sqrt(n); i++) {
                if (n % i == 0) {
                    if (n / i == i) {
                        list.add(i);
                    } else {
                        list.add(i);
                        list.add(n / i);
                    }
                }
            }
            return list;
        }


        //    GCD - Using Euclid theorem.
        private static long gcd(long a, long b) {
            if (b == 0) {
                return a;
            }

            return gcd(b, a % b);
        }

        //    Extended euclidean algorithm
//    Used to solve equations of the form ax + by = gcd(a,b)
//    return array [d, a, b] such that d = gcd(p, q), ap + bq = d
        static long[] extendedEuclidean(long p, long q) {
            if (q == 0)
                return new long[]{p, 1, 0};

            long[] vals = extendedEuclidean(q, p % q);
            long d = vals[0];
            long a = vals[2];
            long b = vals[1] - (p / q) * vals[2];
            return new long[]{d, a, b};
        }

        // X ^ y mod p
        static long power(long x, long y, long p) {
            long res = 1;
            x = x % p;

            while (y > 0) {
                if ((y & 1) == 1)
                    res = (res * x) % p;
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

        // Returns modulo inverse of a
        // with respect to m using extended
        // Euclid Algorithm. Refer below post for details:
        // https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/
        static long inv(long a, long m) {
            long m0 = m, t, q;
            long x0 = 0, x1 = 1;

            if (m == 1)
                return 0;

            // Apply extended Euclid Algorithm
            while (a > 1) {
                q = a / m;
                t = m;
                m = a % m;
                a = t;
                t = x0;
                x0 = x1 - q * x0;
                x1 = t;
            }

            // Make x1 positive
            if (x1 < 0)
                x1 += m0;

            return x1;
        }

        // k is size of num[] and rem[].
        // Returns the smallest number
        // x such that:
        // x % num[0] = rem[0],
        // x % num[1] = rem[1],
        // ..................
        // x % num[k-2] = rem[k-1]
        // Assumption: Numbers in num[] are pairwise
        // coprime (gcd for every pair is 1)
        static long findMinX(long num[], long rem[], long k) {
            int prod = 1;
            for (int i = 0; i < k; i++)
                prod *= num[i];

            int result = 0;
            for (int i = 0; i < k; i++) {
                long pp = prod / num[i];
                result += rem[i] * inv(pp, num[i]) * pp;
            }

            return result % prod;
        }
    }

    private static class BS {
        //  Binary search
        private static int binarySearch(int[] arr, int ele) {
            int low = 0;
            int high = arr.length - 1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (arr[mid] == ele) {
                    return mid;
                } else if (ele < arr[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return -1;
        }

        //    First occurence using binary search
        private static int binarySearchFirstOccurence(int[] arr, int ele) {
            int low = 0;
            int high = arr.length - 1;
            int ans = -1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (arr[mid] == ele) {
                    ans = mid;
                    high = mid - 1;
                } else if (ele < arr[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return ans;
        }


        //    Last occurenece using binary search
        private static int binarySearchLastOccurence(int[] arr, int ele) {
            int low = 0;
            int high = arr.length - 1;
            int ans = -1;
            while (low <= high) {
                int mid = (low + high) / 2;
                if (arr[mid] == ele) {
                    ans = mid;
                    low = mid + 1;
                } else if (ele < arr[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return ans;
        }
    }

    private static class arrays {
        // Merge sort
        static void merge(int arr[], int l, int m, int r) {
            int n1 = m - l + 1;
            int n2 = r - m;
            int L[] = new int[n1];
            int R[] = new int[n2];
            for (int i = 0; i < n1; ++i)
                L[i] = arr[l + i];
            for (int j = 0; j < n2; ++j)
                R[j] = arr[m + 1 + j];
            int i = 0, j = 0;
            int k = l;
            while (i < n1 && j < n2) {
                if (L[i] <= R[j]) {
                    arr[k] = L[i];
                    i++;
                } else {
                    arr[k] = R[j];
                    j++;
                }
                k++;
            }

            while (i < n1) {
                arr[k] = L[i];
                i++;
                k++;
            }

            while (j < n2) {
                arr[k] = R[j];
                j++;
                k++;
            }
        }

        static void sort(int arr[], int l, int r) {
            if (l < r) {
                int m = (l + r) / 2;
                sort(arr, l, m);
                sort(arr, m + 1, r);
                merge(arr, l, m, r);
            }
        }

        static void sort(int[] arr) {
            sort(arr, 0, arr.length - 1);
        }
    }

    private static class UnionFindDisjointSet {
        int[] parent;
        int[] size;
        int n;
        int size1;

        public UnionFindDisjointSet(int n) {
            this.n = n;
            this.parent = new int[n];
            this.size = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }

            for (int i = 0; i < n; i++) {
                size[i] = 1;
            }

            this.size1 = n;
        }

        private int numDisjointSets() {
            System.out.println(size1);
            return size1;
        }

        private boolean find(int a, int b) {
            int rootA = root(a);
            int rootB = root(b);
            if (rootA == rootB) {
                return true;
            }
            return false;
        }

        private int root(int b) {
            while (parent[b] != b) {
                parent[b] = parent[parent[b]];
                b = parent[b];
            }
            return b;
        }

        private void union(int a, int b) {
            int rootA = root(a);
            int rootB = root(b);
            if (rootA == rootB) {
                return;
            }

            if (size[rootA] < size[rootB]) {
                parent[rootA] = parent[rootB];
                size[rootB] += size[rootA];
            } else {
                parent[rootB] = parent[rootA];
                size[rootA] += size[rootB];
            }
            size1--;
            System.out.println(Arrays.toString(parent));
        }
    }

    private static class SegTree {
        int[] st;
        int[] arr;

        public SegTree(int[] arr) {
            this.arr = arr;
            int size = (int) Math.ceil(Math.log(arr.length) / Math.log(2));
            st = new int[(int) ((2 * Math.pow(2, size)) - 1)];
            buildSegmentTree(1, 0, arr.length - 1);
        }

        //**********JUST CALL THE CONSTRUCTOR, THIS FUNCTION WILL BE CALLED AUTOMATICALLY******
        private void buildSegmentTree(int index, int L, int R) {
            if (L == R) {
                st[index] = arr[L];
                return;
            }

            buildSegmentTree(index * 2, L, (L + R) / 2);
            buildSegmentTree(index * 2 + 1, (L + R) / 2 + 1, R);
//          Replace this line if you want to change the function of the Segment tree.
            st[index] = Math.min(st[index * 2], st[index * 2 + 1]);
        }

        //***********We have to use this function **************
        private int Query(int queL, int queR) {
            return Query1(1, 0, arr.length - 1, queL, queR);
        }

        //This is a helper function.
        //************* DO NOT USE THIS ****************
        private int Query1(int index, int segL, int segR, int queL, int queR) {
            if (queL > segR || queR < segL) {
                return -1;
            }

            if (queL <= segL && queR >= segR) {
                return st[index];
            }

            int ans1 = Query1(index * 2, segL, (segL + segR) / 2, queL, queR);
            int ans2 = Query1(index * 2 + 1, (segL + segR) / 2 + 1, segR, queL, queR);
            if (ans1 == -1) {
                return ans2;
            }

            if (ans2 == -1) {
                return ans1;
            }
//             Segment tree implemented for range minimum query. Change the below line to change the function.
            return Math.min(ans1, ans2);
        }

        private void update(int idx, int val) {
            update1(1, 0, arr.length - 1, idx, val);
        }

        private void update1(int node, int queL, int queR, int idx, int val) {
            // idx - index to be updated in the array
            // node - index to be updated in the seg tree
            if (queL == queR) {
                // Leaf node
                arr[idx] += val;
                st[node] += val;
            } else {
                int mid = (queL + queR) / 2;
                if (queL <= idx && idx <= mid) {
                    // If idx is in the left child, recurse on the left child
                    update1(2 * node, queL, mid, idx, val);
                } else {
                    // if idx is in the right child, recurse on the right child
                    update1(2 * node + 1, mid + 1, queR, idx, val);
                }
                // Internal node will have the min of both of its children
                st[node] = Math.min(st[2 * node], st[2 * node + 1]);
            }
        }
    }

    private static class FastReader {
        InputStream is;

        public FastReader(boolean onlineJudge) {
            is = onlineJudge ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        }

        byte[] inbuf = new byte[1024];
        public int lenbuf = 0, ptrbuf = 0;

        int readByte() {
            if (lenbuf == -1)
                throw new InputMismatchException();
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0)
                    return -1;
            }
            return inbuf[ptrbuf++];
        }

        boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }

        int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b))
                ;
            return b;
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        char nextChar() {
            return (char) skip();
        }

        String next() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        String nextLine() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while ((!isSpaceChar(b) || b == ' ')) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        char[] next(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        int nextInt() {
            int num = 0, b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        long nextLong() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        char[][] nextMatrix(int n, int m) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++)
                map[i] = next(m);
            return map;
        }

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        int[][] next2DInt(int n, int m) {
            int[][] arr = new int[n][];
            for (int i = 0; i < n; i++) {
                arr[i] = nextIntArray(m);
            }
            return arr;
        }

        long[][] next2DLong(int n, int m) {
            long[][] arr = new long[n][];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLongArray(m);
            }
            return arr;
        }

        int[] shuffle(int[] arr) {
            Random r = new Random();
            for (int i = 1, j; i < arr.length; i++) {
                j = r.nextInt(i);
                arr[i] = arr[i] ^ arr[j];
                arr[j] = arr[i] ^ arr[j];
                arr[i] = arr[i] ^ arr[j];
            }
            return arr;
        }

        int[] uniq(int[] arr) {
            Arrays.sort(arr);
            int[] rv = new int[arr.length];
            int pos = 0;
            rv[pos++] = arr[0];
            for (int i = 1; i < arr.length; i++) {
                if (arr[i] != arr[i - 1]) {
                    rv[pos++] = arr[i];
                }
            }
            return Arrays.copyOf(rv, pos);
        }
    }

}
