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.*;

public class TaskF {

	public static void main(String[] args) {
		new TaskF().solve();
	}

	private void solve() {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long k = in.nextLong();

		Proger[] raits = new Proger[n];
		for (int i = 0; i < n; i++) {
			int r = in.nextInt();
			raits[i] = new Proger(i, r);
		}

		for (int i = 0; i < k; i++) {
			int x = in.nextInt() - 1;
			int y = in.nextInt() - 1;
			raits[x].addSor(raits[y]);
			raits[y].addSor(raits[x]);
		}

		Arrays.sort(raits, new Comparator<Proger>() {
			@Override
			public int compare(Proger o1, Proger o2) {
				return Integer.compare(o1.getR(), o2.getR());
			}
		});

		for (int i = 0, firstRait = 0; i < n; i++) {
			if (raits[i].getR() != raits[firstRait].getR())
				firstRait = i;

			raits[i].setAns(firstRait);

		}

		Arrays.sort(raits, new Comparator<Proger>() {

			@Override
			public int compare(Proger o1, Proger o2) {
				return Integer.compare(o1.getIndex(), o2.getIndex());
			}
		});

		for (int i = 0; i < n; i++) {
			System.out.println(raits[i].getAns());
		}

	}
}

class Proger {

	private int index, countLessSor, r, ans;

	public Proger(int index, int r) {
		this.index = index;
		this.r = r;
		countLessSor = 0;
		ans = 0;
	}

	void addSor(Proger p) {
		if (p.r < this.r)
			countLessSor++;
	}

	public int getR() {
		return r;
	}

	public int getIndex() {
		return index;
	}

	public void setAns(int iSort) {
		ans = iSort - countLessSor;
	}

	public int getAns() {
		return ans;
	}
}
