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.Scanner;
import java.util.Arrays;
import java.util.Comparator;
public class Main {
    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();
        int arr[][] = new int[n][2];
        int safe[] = new int[n];
        int max = -1;
        for(int i = 0; i < n; i++){
            arr[i][0] = s.nextInt();
            arr[i][1] = i;
            safe[i] = arr[i][0];
            max = Math.max(max,safe[i]);
        }
        int dp[] = new int[n + 1];
        for(int i = 0; i < k; i++){
            int a = s.nextInt();
            int b = s.nextInt();
            if(safe[a - 1] > safe[b - 1]){
                dp[a - 1]--;
            }
            else if(safe[a - 1] < safe[b - 1]){
                dp[b - 1]--;
            }
        }
        
         Arrays.sort(arr,new Comparator<int[]>(){
            public int compare(int a[],int b[]){
                if(a[0] == b[0]){
                    return a[1] - b[1];
                }
         
                return a[0] - b[0];
            }
        });
        int ans[] = new int[n + 1];
        for(int i = 0; i < n; i++){
            int start = 0,end = i,index = -1;
            while(start <= end){
                int mid = (start + end)/2;
                if(arr[mid][0] < arr[i][0]){
                    index = mid;
                    start = mid + 1;
                }
                else{
                    end = mid - 1;
                }
            }
            ans[arr[i][1]] = index + 1;

        }
         for(int i = 0; i < n; i++){
            System.out.print(Math.max(0,ans[i] + dp[i]) + " ");
        }
        
    }
}