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.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Main {

	static InputReader scan = new InputReader(new BufferedInputStream(System.in));
	static ArrayList<Integer> list = new ArrayList<>();
	static Map<Integer, Integer> sum = new HashMap<>();
	static Map<Integer, Integer> map = new HashMap<>();
	static int n, k;

	public static void main(String[] args) {
		n = scan.nextInt();
		k = scan.nextInt();
		for (int i = 0; i < n; i++) {
			int temp = scan.nextInt();
			list.add(temp);
		}
		ArrayList<Integer> listcopy = (ArrayList<Integer>) list.clone();
		for (int i = 0; i < k; i++) {
			int left = scan.nextInt();
			int right = scan.nextInt();
			int value = listcopy.get(left - 1);
			if (value > listcopy.get(right - 1)) {
				if (map.containsKey(left)) {
					map.put(left, map.get(left) + 1);
				} else {
					map.put(left, 1);
				}
			} else if (value < listcopy.get(right - 1)) {
				if (map.containsKey(right)) {
					map.put(right, map.get(right) + 1);
				} else {
					map.put(right, 1);
				}
			}
		}
		Collections.sort(list);
		findsum();
		for (int i = 0; i < listcopy.size(); i++) {
			int number = sum.get(listcopy.get(i));
			if (map.containsKey(i + 1)) {
				number = number - map.get(i + 1);
			}
			System.out.print(number + " ");
		}

	}

	static void findsum() {
		sum.put(list.get(0), 0);
		for (int i = 1; i < list.size(); i++) {
			int left = list.get(i - 1);
			int right = list.get(i);
			if (left < right) {
				sum.put(right, i);
			}
		}
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}
