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.*;
import java.util.*;

public class F {

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		int n = scn.nextInt();
		int k = scn.nextInt();
		Integer a[] = new Integer[n + 1];
		Integer b[] = new Integer[n];
		for (int i = 0; i < n; i++) {
			a[i + 1] = scn.nextInt();
			b[i] = a[i + 1];
		}
		Arrays.sort(b);
		int f[] = new int[200001];
		while (k-- > 0) {
			int u = scn.nextInt();
			int v = scn.nextInt();
			if (a[u] > a[v])
				f[u]++;
			if (a[v] > a[u])
				f[v]++;
		}

		for (int i = 1; i <= n; i++) {
			int idx = first(b, 0, n - 1, a[i], n);
			if (idx > 0) {
				int noe = idx;
				noe -= f[i];
				System.out.print(noe + " ");
			} else {
				idx = Arrays.binarySearch(b, a[i] - 1);
				if (idx > 0) {
					while (b[idx] == a[i] - 1)
						idx++;
				} else
					idx = -(idx + 1);
				int noe = idx;
				noe -= f[i];
				System.out.print(noe + " ");
			}
		}
	}

	public static int first(Integer arr[], int low, int high, long x, int n) {
		if (high >= low) {
			int mid = low + (high - low) / 2;
			if ((mid == 0 || x > arr[mid - 1]) && arr[mid] == x)
				return mid;
			else if (x > arr[mid])
				return first(arr, (mid + 1), high, x, n);
			else
				return first(arr, low, (mid - 1), x, n);
		}
		return -1;
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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