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 Main {
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        int arr[][]=new int[n][2];
        for(int i=0;i<n;i++) {
            arr[i][0]=sc.nextInt();
            arr[i][1]=i;
        }
        int a,b;
        int greater[]=new int[n];
        for(int i=0;i<k;i++) {
            a=sc.nextInt();
            b=sc.nextInt();
            if(arr[a-1][0]>arr[b-1][0]) {
                greater[a-1]++;
            }
            
            else if(arr[b-1][0]>arr[a-1][0]) {
                greater[b-1]++;
            }
            
        }
        
        
        java.util.Arrays.sort(arr, new java.util.Comparator<int[]>() {
        public int compare(int[] a, int[] b) {
        return Integer.compare(a[0], b[0]);
        }
        });
        
        int ans[]=new int[n];
        int count=0;
        int last=arr[0][0];
        for(int i=1;i<n;i++) {
            if(arr[i][0]>last) {
                count=i;
                last=arr[i][0];
            }
            ans[arr[i][1]]=count-greater[arr[i][1]];
            //System.out.println(count);
        }
        for(int i=0;i<n;i++) {
            System.out.print(ans[i]+" ");
        }
        System.out.println("");
        /*
        for(int i=0;i<n;i++) {
            System.out.print(arr[i][0]+" ");
        }
        System.out.println("");
        for(int i=0;i<n;i++) {
            System.out.print(greater[i]+" ");
        }
        System.out.println("");
        */

    }
}