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

public class Solution2 {

    private void solve() throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        int[] r = new int[n];
        List<Pair> l = new ArrayList<>(n + 2);
        for (int i = 0; i < n; i++) {
            r[i] = in.nextInt();
            l.add(new Pair(i, r[i]));
        }
        l.add(new Pair(-1, 0));
        l.add(new Pair(-1, 2_000_000_001));
        Collections.sort(l);

        int[] substr = new int[n];
        for (int i = 0; i < k; i++) {
            int x = in.nextInt() - 1;
            int y = in.nextInt() - 1;
            if (r[x] > r[y]) substr[x]++;
            if (r[y] > r[x]) substr[y]++;
        }
        List<Pair> ans = new ArrayList<>(n);
        for (int i = 1; i < n + 1; i++) {
            int count = binarySearch(l, l.get(i).value);
            ans.add(new Pair(count - substr[l.get(i).index], l.get(i).index));
        }
        Collections.sort(ans);
        StringBuilder st = new StringBuilder("");
        for (int i = 0; i < ans.size(); i++) {
            st.append(ans.get(i).index);
            st.append(' ');
        }
        System.out.println(st);
    }

    private int binarySearch(List<Pair> a, long x) {
        int min = 0;
        int max = a.size();
        while (min < max - 1) {
            int mid = min + (max - min) / 2;
            int value = a.get(mid).value;
            if (value < x) {
                min = mid;
            }
            else {
                max = mid;
            }
        }
        return min;
    }

    private class Pair implements Comparable<Pair> {
        int index;
        int value;

        @Override
        public int compareTo(Pair o) {
            return Integer.compare(this.value, o.value);
        }

        public Pair(int i, int v) {
            this.index = i;
            this.value = v;
        }
    }


    private PrintWriter out;
    private MyScanner in;

    private void run() throws IOException {
        in = new MyScanner();
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

    private class MyScanner {
        private BufferedReader br;
        private StringTokenizer st;

        public MyScanner() throws IOException {
            this.br = new BufferedReader(new InputStreamReader(System.in));
        }

        public MyScanner(String fileTitle) throws IOException {
            this.br = new BufferedReader(new FileReader(fileTitle));
        }

        public String nextLine() throws IOException {
            String s = br.readLine();
            return s == null ? "-1" : s;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                String s = br.readLine();
                if (s == null) {
                    return "-1";
                }
                st = new StringTokenizer(s);
            }
            return st.nextToken();
        }

        public Integer nextInt() throws IOException {
            return Integer.parseInt(this.next());
        }

        public Long nextLong() throws IOException {
            return Long.parseLong(this.next());
        }

        public Double nextDouble() throws IOException {
            return Double.parseDouble(this.next());
        }

        public void close() throws IOException {
            this.br.close();
        }
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new Solution2().run();
    }
}




