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.Arrays;
import java.util.Scanner;

public class D {
    static class Programmer implements Comparable {
        final int index;
        final int rank;

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

        @Override
        public int compareTo(Object o) {
            Programmer p = (Programmer) o;
            return p.rank - rank;
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();

        Programmer[] p = new Programmer[n];
        int[] r = new int[n];
        for (int i = 0; i < n; ++i) {
            r[i] = scanner.nextInt();
            p[i] = new Programmer(i, r[i]);
        }

        int[] x = new int[k];
        int[] y = new int[k];
        for (int i = 0; i < k; ++i) {
            x[i] = scanner.nextInt() - 1;
            y[i] = scanner.nextInt() - 1;
        }

        Arrays.sort(p);
        int[] answer = new int[n];
        int start = 0;
        for (int i = 0; i < n; ++i) {
            if (i == n - 1 || p[i].rank != p[i + 1].rank) {
                for (int j = start; j <= i; ++j) {
                    answer[p[j].index] = n - i - 1;
                }
                start = i + 1;
            }
        }

        for (int i = 0; i < k; ++i) {
            if (r[x[i]] > r[y[i]]) {
                --answer[x[i]];
            }
            if (r[y[i]] > r[x[i]]) {
                --answer[y[i]];
            }
        }

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