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.List;
import java.util.Scanner;

public class Main {

	private static class Element implements Comparable<Element> {
		public int value;
		public int index;
		public Element(int value , int index) {
			this.value = value;
			this.index = index;
		}
		@Override		
		public int compareTo(Element e) {
			if (this.value < e.value) {
				return - 1;
			} else if (this.value > e.value) {
				return 1;
			} else {
				return 0;
			}
		}
	}
	
	private static int[] ans = new int[200010];
	private static int[] values = new int[200010];
	
	public static void main(String[] args) throws Exception {	
		
		Scanner scan = new Scanner(System.in);
	
		int i , n , k , sum = 0;
		n = scan.nextInt();
		k = scan.nextInt();
		List<Element> list = new ArrayList<>();
		for (i = 1;i <= n;i ++) {
			int value = scan.nextInt();
			values[i] = value;
			list.add(new Element(value , i));
		}
		Collections.sort(list);
		for (i = 0;i < list.size();i ++) {
			int value = list.get(i).value;
			int temp = 0;	
			while (i < list.size() && (list.get(i).value == value)) {
				ans[list.get(i).index] = sum;
				i ++;
				temp ++;
			}
			i --;
			sum += temp;
		}
		for (i = 0;i < k;i ++) {
			int from , to;
			from = scan.nextInt();
			to = scan.nextInt();
			if (values[from] > values[to]) {
				ans[from] --;
			} else if (values[from] < values[to]) {
				ans[to] --;
			}
		}
		for (i = 1;i <= n;i ++) {
			if (i > 1) {
				System.out.print(" ");
			}
			System.out.print(ans[i]);
		}
		System.out.println();
		
	}
    
}













