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.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class F978 {
	
	class Pair implements Comparable<Pair> {
		int val, index;
		Pair(int v, int i) {
			val = v; index = i;
		}
		public int compareTo(Pair p) {
			if(val == p.val) return index - p.index;
			return val - p.val;
		}
	}
	
	public void solve(Scanner in, PrintWriter out) {
		int n = in.nextInt();
		int m = in.nextInt();
		Pair[] a = new Pair[n];
		for(int i = 0; i < n; ++i) a[i] = new Pair(in.nextInt(), i);
		int[] delta = new int[n];
		for(int i = 0; i < m; ++i) {
			int u = in.nextInt() - 1;
			int v = in.nextInt() - 1;
			if(a[u].val > a[v].val) --delta[u];
			else if(a[v].val > a[u].val) --delta[v];
		}
		Arrays.sort(a);
		int prev = a[0].val;
		int prevAns = 0;
		for(int i = 0; i < n; ++i) {
			if(a[i].val > prev) {
				delta[a[i].index] += i;
				prev = a[i].val;
				prevAns = i;
			} else {
				delta[a[i].index] += prevAns;
			}
		}
		for(int i = 0; i < n; ++i) out.print(delta[i] + " ");
		out.println();
	}
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		new F978().solve(in, out);
		in.close();
		out.close();
	}
}
