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.*;
import java.io.*;

public class A {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		TreeSet<Integer> set = new TreeSet<>();
		int[] r = new int[n];
		for (int i = 0; i < n; i++) {
			r[i] = sc.nextInt();
			set.add(r[i]);
		}
		HashMap<Integer, Integer> map = new HashMap<>();
		int ll = 1;
		for (int x : set)
			map.put(x, ll++);
		for (int i = 0; i < n; i++)
			r[i] = map.get(r[i]);
		int cnt[] = new int[1000001];
		for (int i = 0; i < n; i++)
			cnt[r[i]]++;
		for (int i = 1; i < cnt.length; i++)
			cnt[i] += cnt[i - 1];
		int[] qua = new int[n];
		for (int i = 0; i < k; i++) {
			int x = sc.nextInt()-1;
			int y = sc.nextInt()-1;
			if (r[x] < r[y])
				qua[y]++;
			if (r[y] < r[x])
				qua[x]++;
		}
//		System.out.println(Arrays.toString(cnt));
		PrintWriter pw = new PrintWriter(System.out);
		for (int i = 0; i < n; i++)
			pw.print(cnt[r[i]-1] - qua[i] + " ");
		pw.flush();
	}

	static class Event implements Comparable<Event> {
		int sub, exam;

		public Event(int a, int b) {
			sub = a;
			exam = b;
		}

		@Override
		public int compareTo(Event o) {
			if (exam != o.exam)
				return exam - o.exam;
			return sub - o.sub;
		}
	}

	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());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}
