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.HashMap;
import java.util.Scanner;

public class CF981F {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int nProgs = sc.nextInt();
		int nQuarrels = sc.nextInt();
		ArrayList<Programmer> progs = new ArrayList<>();
		for(int i=0; i<nProgs; i++) {
			progs.add(new Programmer(sc.nextInt(), i));
		}
		for(int i=0; i<nQuarrels; i++) {
			int src = sc.nextInt()-1;
			int dest = sc.nextInt()-1;
			if(progs.get(src).skill>progs.get(dest).skill) progs.get(src).inQuarrel.put(dest, true);
			if(progs.get(dest).skill>progs.get(src).skill) progs.get(dest).inQuarrel.put(src, true);
			
		}
		HashMap<Integer, Integer> startIndices = new HashMap<>();
		int[] ans = new int[nProgs];
		Collections.sort(progs);
		for(int i=0; i<nProgs; i++)
			if(!startIndices.containsKey(progs.get(i).skill)) startIndices.put(progs.get(i).skill, i);
		for(int i=0; i<nProgs; i++)
			ans[progs.get(i).o_index] = startIndices.get(progs.get(i).skill)-progs.get(i).inQuarrel.size();
		StringBuilder sb = new StringBuilder();
		for(int i=0; i<nProgs; i++) sb.append(ans[i]).append(" ");
		System.out.println(sb.toString().trim());
	}
	
	static class Programmer implements Comparable<Programmer> {
		HashMap<Integer, Boolean> inQuarrel;
		int skill;
		int o_index;
		
		public Programmer(int skill, int o_index) {
			this.skill = skill;
			this.o_index = o_index;
			inQuarrel = new HashMap<>();
		}

		@Override
		public int compareTo(Programmer o) {
			return Integer.compare(this.skill, o.skill);
		}
	}
}
