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

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

        String[] raw = in.readLine().split(" ");

        int n = Integer.parseInt(raw[0]);
        int k = Integer.parseInt(raw[1]);

        raw = in.readLine().split(" ");

        TreeMap<Integer, Integer> skills = new TreeMap<>();
        int[] r = new int[n];
        for (int i = 0; i < n; i++) {
            r[i] = Integer.parseInt(raw[i]);

            if (!skills.containsKey(r[i])) {
                skills.put(r[i], 0);
            }

            skills.put(r[i], skills.get(r[i]) + 1);
        }

        int lessThen = 0;
        Map<Integer, Integer> lessSkills = new TreeMap<>();
        Set<Integer> keys = skills.keySet();
        for (Integer key : keys) {
            lessSkills.put(key, lessThen);
            lessThen += skills.get(key);
        }

        ArrayList<Integer> quarrels = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            quarrels.add(0);
        }

        for (int i = 0; i < k; i++) {
            raw = in.readLine().split(" ");

            int a = Integer.parseInt(raw[0]) - 1;
            int b = Integer.parseInt(raw[1]) - 1;

            if (r[a] > r[b])
                quarrels.set(a, quarrels.get(a) + 1);

            if (r[b] > r[a])
                quarrels.set(b, quarrels.get(b) + 1);
        }

        for (int i = 0; i < n; i++) {
            int lessCnt = lessSkills.getOrDefault(r[i], 0);
            int quarrelCnt = quarrels.get(i);

            System.out.printf("%d ", lessCnt - quarrelCnt);
        }
    }
}