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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;

public class Mentors {

	public static void main(String [] args) throws IOException {
		Bersoft bs = new Bersoft();
		bs.solve();
	}
	
}

class Bersoft {
	
	Bersoft() throws IOException {
		QuickInput in = new QuickInput();
		n = in.nextInt();
		k = in.nextInt();
		r = new int [n];
		for (int y = 0; y < n; ++y)
			r[y] = in.nextInt();
		
		conflict = new int [n];
		Arrays.fill(conflict, 0);
		while(k-- > 0) {
			int a = in.nextInt() - 1;
			int b = in.nextInt() - 1;
			conflict[b] += (r[a] < r[b]) ? 1 : 0;
			conflict[a] += (r[a] > r[b]) ? 1 : 0;
		}
	}
	
	void solve() {
		Pair [] prog = new Pair [n];
		for (int v = 0; v < n; ++v)
			prog[v] = new Pair(v, r[v]);
		Arrays.sort(prog, (Pair a, Pair b) -> {
			if (a.r < b.r)
				return -1;
			if (a.r > b.r)
				return 1;
			return 0;
		});
		
		int [] ans = new int [n];
		int lvl = 0, amount = 0;
		
		for (int i = 0; i < n; ++i) {
			int j = prog[i].index;
			if (lvl != prog[i].r) {
				lvl = prog[i].r;
				amount = 0;
			}
			ans[j] = i - amount - conflict[j];
			++amount;
		}
		
		PrintWriter out = new PrintWriter(System.out);
		for (int num : ans)
			out.print(num + " ");
		out.flush();
		out.close();
	}
	
	int n, k;
	int [] r;
	int [] conflict;
	
}

class Pair {
	
	int index, r;
	
	Pair(int index, int r){
		this.index = index;
		this.r = r;
	}
	
	public String toString() {
		return "(" + index + ", " + r + ")";
	}
	
}

class QuickInput {
	
	StreamTokenizer stk;
	
	QuickInput() {
		stk = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	}
	
	int nextInt() throws IOException {
		stk.nextToken();
		return (int) stk.nval;
	}
	
}