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 F {
	
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt(), k = sc.nextInt();
		
		ArrayList<Integer> qua[] = new ArrayList[n+1];
		
		PriorityQueue<Node> pq = new PriorityQueue<>(new Comparator<Node>(){

			@Override
			public int compare(Node a, Node b) {
				return b.skill-a.skill;
			};
		
		});
		
		for(int i=1;i<=n;i++){
			pq.offer(new Node(i, sc.nextInt()));
			qua[i] = new ArrayList<>();
		}
		for(int i=1;i<=k;i++){
			int x = sc.nextInt(), y = sc.nextInt();
			qua[x].add(y);
			qua[y].add(x);
		}
		
		boolean vis[] = new boolean[n+1];
		int ans[] = new int[n+1];
		
		while(pq.size()>0){
			List<Integer> tmp = new ArrayList<>();
			int sk = pq.peek().skill;
			while(pq.size() > 0 && pq.peek().skill == sk){
				Node cur = pq.poll();
				vis[cur.id] = true;
				tmp.add(cur.id);
			}
			
			for(int i:tmp){
				int cnt = 0;
				for(int j:qua[i]){
					if(!vis[j])
						cnt++;
				}
				ans[i] = pq.size() - cnt;
			}
		}
		
		StringBuilder sb = new StringBuilder();
		for(int i=1;i<=n;i++){
			sb.append(ans[i]+" ");
		}
		System.out.println(sb);
		
		
	}
}
class Node{
	int id;
	int skill;
	Node(int i, int s){
		id = i;
		skill = s;
	}
}
