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 {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Solution solver = new Solution();
        solver.solve(in, out);
        out.close();
    }

    static class Solution {

        public void solve(InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            Programmer[] p = new Programmer[n];
            for(int i = 0; i < n; ++i) {
                p[i] = new Programmer(i, in.nextInt());
            }
            for(int i = 0; i < k; ++i) {
                int x = in.nextInt() - 1;
                int y = in.nextInt() - 1;
                if(p[x].skill == p[y].skill) continue;
                p[x].quarrel.add(y);
                p[y].quarrel.add(x);
            }
            Arrays.sort(p);
            HashSet<Integer> runningSet = new HashSet<>();
            HashMap<Integer, Integer> seen = new HashMap<>();
            int[] res = new int[n];
            for(int i = 0; i < n; ++i) {
                p[i].quarrel.retainAll(runningSet);
                res[p[i].index] = Math.max(0, i - p[i].quarrel.size() - seen.getOrDefault(p[i].skill, 0));
                runningSet.add(p[i].index);
                seen.put(p[i].skill, seen.getOrDefault(p[i].skill, 0) + 1);
            }
            for(int i = 0; i < n; ++i) {
                out.printf("%d ", res[i]);
            }
        }
    }

    static class Programmer implements Comparable<Programmer> {
        int index;
        int skill;
        HashSet<Integer> quarrel;

        public Programmer(int index, int skill) {
            this.index = index;
            this.skill = skill;
            quarrel = new HashSet<>();
        }

        public int compareTo(Programmer p) {
            return Integer.compare(skill, p.skill);
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

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

        String nextLine() {
            String s = "";
            try {
                s = reader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }
    }
}