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.StringTokenizer;
import java.util.TreeMap;

public class CF978F {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer s = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(s.nextToken());
        int k = Integer.parseInt(s.nextToken());
        int [] skills = new int [n];
        s = new StringTokenizer(br.readLine());
        TreeMap<Integer,Integer> orderedSkills = new TreeMap<>();
        for (int i = 0; i < n; i++) {
            skills[i] = Integer.parseInt(s.nextToken());
            if (orderedSkills.containsKey(skills[i])) orderedSkills.put(skills[i],orderedSkills.get(skills[i])+1);
            else orderedSkills.put(skills[i],1);
        }
        TreeMap<Integer,Integer> numPairs = new TreeMap<>(); // number of pairs you can make based on skill only, contains (skill, # of ppl that person can mentor)
        int count = 0;
        for (int skill : orderedSkills.keySet()) {
            numPairs.put(skill,count);
            count += orderedSkills.get(skill);
        }
        int [] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = numPairs.get(skills[i]);
        }
        for (int i = 0; i < k; i++) {
            s = new StringTokenizer(br.readLine());
            int x = Integer.parseInt(s.nextToken());
            int y = Integer.parseInt(s.nextToken());
            if (skills[x-1]>skills[y-1]) {
                res[x-1]--;
            }
            else if (skills[y-1] > skills[x-1]){
                res[y-1]--;
            }
        }
        for (int num : res) {
            System.out.print(num+" ");
        }
    }
}
