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.*;

public class Solution {
    public static void main(String[] args) {
        new Solution().solve();
    }

    public void solve() {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        List<Pair> l = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            l.add(new Pair(in.nextInt(), i, 0));
        }
        Collections.sort(l, new PairComp());
        for (int i = 0; i < n; i++) {
            if (i != 0 && l.get(i - 1).s == l.get(i).s) {
                l.get(i).mentees = l.get(i - 1).mentees;
            } else {
                l.get(i).mentees = i;
            }
        }
        Collections.sort(l, new InComp());
        for (int i = 0; i < k; i++) {
            int a = in.nextInt() - 1;
            int b = in.nextInt() - 1;
            if (l.get(a).s < l.get(b).s) {
                l.get(b).mentees--;
            } else if (l.get(a).s > l.get(b).s) {
                l.get(a).mentees--;
            }
        }
        for (int i = 0; i < n; i++) {
            System.out.printf("%d ", l.get(i).mentees);
        }
    }

    class Pair {
        int s;
        int i;
        int mentees;

        public Pair(int s, int i, int mentees) {
            this.s = s;
            this.i = i;
            this.mentees = mentees;
        }
    }

    class PairComp implements Comparator<Pair> {
        @Override
        public int compare(Pair o1, Pair o2) {
            return Integer.compare(o1.s, o2.s);
        }
    }

    class InComp implements Comparator<Pair> {
        @Override
        public int compare(Pair o1, Pair o2) {
            return Integer.compare(o1.i, o2.i);
        }
    }
}