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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF978F extends PrintWriter {
	CF978F() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF978F o = new CF978F(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] aa = new int[n];
		Integer[] ii = new Integer[n];
		for (int i = 0; i < n; i++) {
			aa[i] = sc.nextInt();
			ii[i] = i;
		}
		Arrays.sort(ii, (i, j) -> aa[i] - aa[j]);
		int[] kk = new int[n];
		for (int i = 0, j; i < n; i = j) {
			int a = aa[ii[i]];
			for (j = i; j < n && aa[ii[j]] == a; j++)
				kk[ii[j]] = i;
		}
		while (k-- > 0) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			if (aa[i] < aa[j])
				kk[j]--;
			else if (aa[i] > aa[j])
				kk[i]--;
		}
		for (int i = 0; i < n; i++)
			print(kk[i] + " ");
		println();
	}
}
