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.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;

public class mentors {

    static class Point implements Comparable<Point>{
        int ind;
        int v;
        Point(int v, int ind){
            this.v = v;
            this.ind = ind;
        }
        public int compareTo(Point s){
            return v - s.v;
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken()), k = Integer.parseInt(st.nextToken());
        Point[] ar = new Point[n];
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            ar[i] = new Point(Integer.parseInt(st.nextToken()), i);
        }
        int[] ans = new int[n];
        for (int i = 0; i < k; i++) {
            st = new StringTokenizer(br.readLine());
            int a = Integer.parseInt(st.nextToken()) - 1, b = Integer.parseInt(st.nextToken()) - 1;
            if(ar[a].v > ar[b].v){
                ans[a]--;
            }else if(ar[b].v > ar[a].v){
                ans[b]--;
            }
        }
        Arrays.sort(ar);
        int pointer = 0;
        ArrayList<Integer> a = new ArrayList<Integer>();
        for (int i = n - 1; i >= 0; i--) {
              if(i > 0 && ar[i].v == ar[i - 1].v){
                  a.add(ar[i].ind);
              }else{
                  a.add(ar[i].ind);
                  for (int j = 0; j < a.size(); j++) {
                      ans[a.get(j)] += i;
                  }
                  a.clear();
              }
        }
        for (int i = 0; i < n; i++) {
            System.out.print(ans[i] + " ");
        }
    }
}
