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.ArrayList;
import java.util.Collections;
import java.util.Scanner;



public class Main {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt(), k = s.nextInt();
		int[] temp = new int[n];
		ArrayList<node> arr = new ArrayList<>();
		for(int i = 0; i < n; i++){
			node t = new node();
			t.idx = i;
			temp[i] = t.value = s.nextInt();
			arr.add(t);
		}
		Collections.sort(arr);
		int[] ans = new int[n];
		for(int i = 0; i < arr.size(); ){
			node t = arr.get(i);
			int count = 1;
			for(int j = i + 1;j < arr.size(); j++){
				if(t.value == arr.get(j).value){
					count++;
				}else{
					break;
				}
			}
			ans[t.idx] = n - i - count;
			int c = count;
			while(--count>0){
				ans[arr.get(i + count).idx] = ans[t.idx];
			}
			i += c;
		}
		for(int i = 0; i < k; i++){
			int a = s.nextInt() - 1, b = s.nextInt() - 1;
			if(temp[a] > temp[b]){
				ans[a]--;
			}else if(temp[a] < temp[b]){
				ans[b]--;
			}
		}
		for(int i = 0; i < ans.length; i++){
			System.out.print(ans[i] + " ");
		}

	}

}


class node implements Comparable<node>{
	int idx, value;

	@Override
	public int compareTo(node arg0) {
		return -this.value + arg0.value;
	}
}