In BerSoft n programmers work, the programmer i is characterized by a skill r_i.

A programmer a can be a mentor of a programmer b if and only if the skill of the programmer a is strictly greater than the skill of the programmer b (r_a > r_b) and programmers a and b are not in a quarrel.

You are given the skills of each programmers and a list of k pairs of the programmers, which are in a quarrel (pairs are unordered). For each programmer i, find the number of programmers, for which the programmer i can be a mentor.

Input

The first line contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k ≤ min(2 ⋅ 10^5, (n ⋅ (n - 1))/(2))) — total number of programmers and number of pairs of programmers which are in a quarrel.

The second line contains a sequence of integers r_1, r_2, ..., r_n (1 ≤ r_i ≤ 10^{9}), where r_i equals to the skill of the i-th programmer.

Each of the following k lines contains two distinct integers x, y (1 ≤ x, y ≤ n, x ≠ y) — pair of programmers in a quarrel. The pairs are unordered, it means that if x is in a quarrel with y then y is in a quarrel with x. Guaranteed, that for each pair (x, y) there are no other pairs (x, y) and (y, x) in the input.

Output

Print n integers, the i-th number should be equal to the number of programmers, for which the i-th programmer can be a mentor. Programmers are numbered in the same order that their skills are given in the input.

Examples

Input

4 2
10 4 10 15
1 2
4 3


Output

0 0 1 2 


Input

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5


Output

5 4 0 5 3 3 9 0 2 5 

Note

In the first example, the first programmer can not be mentor of any other (because only the second programmer has a skill, lower than first programmer skill, but they are in a quarrel). The second programmer can not be mentor of any other programmer, because his skill is minimal among others. The third programmer can be a mentor of the second programmer. The fourth programmer can be a mentor of the first and of the second programmers. He can not be a mentor of the third programmer, because they are in a quarrel.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Pranay2516
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        FMentors solver = new FMentors();
        solver.solve(1, in, out);
        out.close();
    }

    static class FMentors {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int n = in.nextInt(), k = in.nextInt();
            int[] r = in.readArray(n);
            Pair<Integer, Integer>[] a = new Pair[n];
            for (int i = 0; i < n; ++i) a[i] = new Pair<>(r[i], i);
            func.sort(a);
            int[] cnt = new int[n];
            int prev = -1, count = 0;
            for (int i = 0; i < n; ++i) {
                if (prev == -1 || prev == a[i].x) cnt[a[i].y] = count;
                else cnt[a[i].y] = i;
                prev = a[i].x;
                count = cnt[a[i].y];
            }
            while (k-- > 0) {
                int u = in.nextInt() - 1, v = in.nextInt() - 1;
                if (r[u] > r[v]) cnt[u]--;
                else if (r[v] > r[u]) cnt[v]--;
            }
            for (int i = 0; i < n; ++i) {
                out.print(cnt[i] + " ");
            }
            out.println();
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private FastReader.SpaceCharFilter filter;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1) throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) return -1;
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public int[] readArray(int size) {
            int[] array = new int[size];
            for (int i = 0; i < size; i++) array[i] = nextInt();
            return array;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }

    static class Pair<U, V> implements Comparable<Pair<U, V>> {
        public U x;
        public V y;

        public Pair(U x, V y) {
            this.x = x;
            this.y = y;
        }

        public int compareTo(Pair<U, V> o) {
            int value = ((Comparable<U>) x).compareTo(o.x);
            if (value != 0) return value;
            return ((Comparable<V>) y).compareTo(o.y);
        }

    }

    static class func {
        public static <T extends Comparable<? super T>> void sort(T[] arr) {
            //Taken from @KharYusuf
            int n = arr.length, mid, h, s, l, i, j, k;
            T[] res = arr.clone();
            n--;
            for (s = 1; s <= n; s <<= 1) {
                for (l = 0; l < n; l += (s << 1)) {
                    h = Math.min(l + (s << 1) - 1, n);
                    mid = Math.min(l + s - 1, n);
                    i = l;
                    j = mid + 1;
                    k = l;
                    while (i <= mid && j <= h)
                        res[k++] = arr[i].compareTo(arr[j]) <= 0 ? arr[i++] : arr[j++];
                    while (i <= mid) res[k++] = arr[i++];
                    while (j <= h) res[k++] = arr[j++];
                    for (k = l; k <= h; k++) arr[k] = res[k];
                }
            }
        }

    }
}

