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.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class CF978F {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int n = in.nextInt();
        int k = in.nextInt();
        List<Programmer> programmers = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int skill = in.nextInt();
            programmers.add(new Programmer(skill, i));
        }
        for (int i = 0; i < k; i++) {
            int a = in.nextInt() - 1;
            int b = in.nextInt() - 1;
            if (programmers.get(a).skill > programmers.get(b).skill) {
                programmers.get(a).hates++;

            } else if (programmers.get(b).skill > programmers.get(a).skill) {
                programmers.get(b).hates++;
            }
        }
        int[] mentors = new int[n];
        Collections.sort(programmers, Comparator.comparingInt(a -> a.skill));
        for (int i = 0; i < n; i++) {
            Programmer programmerA = programmers.get(i);
            int lb = lowerBound(programmers, programmerA.skill);
            mentors[programmerA.index] = lb - programmerA.hates + 1;

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

    public static int lowerBound(List<Programmer> programmers, int value) {
        int low = 0;
        int high = programmers.size();
        while (low < high) {
            final int mid = (low + high) / 2;
            if (value <= programmers.get(mid).skill) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low - 1;
    }

    public static class Programmer {
        int skill;
        int index;
        int hates = 0;

        public Programmer(int skill, int index) {
            this.skill = skill;
            this.index = index;
        }
    }


}
