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.*;
import java.util.*;

public class F {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    public static void main(String[] args) throws IOException {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter(new File("output.txt"));
        } catch (Exception e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
        }

        tok = new StringTokenizer("");
        new F().solve();
        out.flush();
        out.close();
    }

    String nextTok() throws IOException {
        if (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextTok());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextTok());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextTok());
    }

    static class Point implements Comparable<Point> {

        int x;
        int y;

        public Point( int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public int compareTo(Point o) {

            if (o.x < x) return 1;
            if (o.x > x) return -1;
            if (o.y > y) return -1;
            if (o.y < y) return 1;
            return 0;
        }
    }

    static class Pair implements Comparable<Pair> {

        int ind;
        int val;

        public Pair( int ind, int val) {
            this.ind = ind;
            this.val = val;
        }

        @Override
        public int compareTo(Pair o) {

            if (o.val > val) return 1;
            if (o.val < val) return -1;
            if (o.ind > ind) return -1;
            if (o.ind < ind) return 1;
            return 0;
        }

    }

    void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
//        int r[] = new int[n];
        int ans[] = new int[n+1];
        TreeSet<Point> ts = new TreeSet<>();
        Pair[] r = new Pair[n];
        for (int i = 0; i < n; i++) {
            r[i] = new Pair(i+1, nextInt());
        }

        for (int i = 0; i < k; i++) {
            int v = nextInt();
            int u = nextInt();
            if (r[v-1].val < r[u-1].val) {
                ans[u]--;
            } else if (r[u-1].val < r[v-1].val) {
                ans[v]--;
            }
        }
        Arrays.sort(r);


        for (int i = 0; i < n; i++) {
            int nextInd = -1;
            long l = i+1;
            long rr = n-1;
            while (l <= rr) {
                long mid = (l+rr)/2;
                if (r[(int)mid].val < r[i].val) {
                    rr = mid - 1;
                    nextInd = (int)mid;
                } else {
                    l = mid + 1;
                }
            }
            if (nextInd != -1) {
                ans[r[i].ind] += n - nextInd;
            } else {
                ans[r[i].ind] = 0;
            }
        }

        for (int i = 0; i < n; i++) {
            out.print(ans[i+1] + " ");
        }


    }



}