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.*;

/**
 * @author radium4ye
 */
public class Test {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();

        int[] result = new int[n];
        int[] quarrelResult = new int[n];

        //技能
        List<Skill> list = new ArrayList<>(n);
        for (int i = 0; i < n; i++) {
            Skill skill = new Skill(s.nextInt(), i);
            list.add(skill);
        }

        //争吵
        for (int i = 0; i < k; i++) {
            int x = s.nextInt() - 1;
            int y = s.nextInt() - 1;
            Skill skillx = list.get(x);
            Skill skilly = list.get(y);

            if (skillx.value < skilly.value) {
                quarrelResult[y]--;
            } else if (skilly.value < skillx.value) {
                quarrelResult[x]--;
            }

        }

        list.sort(Comparator.comparing(o -> o.value));

        int preValue = list.get(0).value;
        int preResult = 0;
        for (int i = 1; i < n; i++) {
            Skill skill = list.get(i);

            if (preValue == skill.value) {

            } else {
                preValue = skill.value;
                preResult = i;
            }

            result[skill.position] = preResult;
        }

        //输出结果
        for (int i = 0; i < n; i++) {
            System.out.printf((result[i] + quarrelResult[i]) + " ");
        }
    }

    public static class Skill {
        Integer value;
        Integer position;

        public Skill(Integer value, Integer position) {
            this.value = value;
            this.position = position;
        }
    }
}
