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.Comparator;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		PriorityQueue<pair> q=new PriorityQueue<>(new comp());
		int n=s.nextInt();
		int k=s.nextInt();
		HashMap<Integer,Integer> map2=new HashMap<>();
		int[] arr=new int[n];
		for(int i=0;i<n;i++) {
			arr[i]=s.nextInt();
			pair ob=new pair(i+1,arr[i]);
			if(map2.containsKey(arr[i])) {
				map2.put(arr[i], map2.get(arr[i])+1);
			}else {
				map2.put(arr[i], 1);
			}
			q.add(ob);
		}
		
		HashMap<Integer,ArrayList<Integer>> map=new HashMap<>();
		
		for(int i=0;i<k;i++) {
			int a=s.nextInt();
			int b=s.nextInt();
			
			if(map.containsKey(a)) {
				ArrayList<Integer> p=map.get(a);
				p.add(b);
				map.put(a, p);
			}else {
				ArrayList<Integer> list=new ArrayList<>();
				list.add(b);
				map.put(a, list);
			}
			
			if(map.containsKey(b)) {
				ArrayList<Integer> p=map.get(b);
				p.add(a);
				map.put(b, p);
			}else {
				ArrayList<Integer> list=new ArrayList<>();
				list.add(a);
				map.put(b, list);
			}
		}
		
		
		int[] ans=new int[n];
		while(!q.isEmpty()) {
			pair ob=q.poll();
			int b=0;
			if(map.containsKey(ob.num)) {
			
				for(int i=0;i<map.get(ob.num).size();i++) {
					if(arr[map.get(ob.num).get(i)-1]<ob.val) {
						b++;
					}
				}
			}
			
			ans[ob.num-1]=q.size()-b-map2.get(ob.val)+1;
			
			map2.put(ob.val, map2.get(ob.val)-1);
			
		}
		
		
		for(int i=0;i<n;i++) {
			System.out.print(ans[i]+" ");
		}
		
		
		
		
		
	}
	
}
class pair{
	int num;
	int val;
	public pair(int num,int val) {
		this.num=num;
		this.val=val;
	}
}
class comp implements Comparator<pair>{
	public int compare(pair h,pair j) {
		if(h.val<j.val) {
			return 1;
		}else if(h.val==j.val){
			return 0;
		}else {
			return -1;
		}
	}
}