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.InputStream;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

	public static void main(String[] args) throws IOException {
		MyScanner sc = new MyScanner(System.in);
		final int n = sc.nextInt();
		final int k = sc.nextInt();
		int[] r = new int[n];
		for(int i=0; i<n; i++) {
			r[i] = sc.nextInt();
		}
		int[] x = new int[k];
		int[] y = new int[k];
		int[] penal = new int[n];
		for(int i=0; i<k; i++) {
			x[i] = sc.nextInt()-1; y[i] = sc.nextInt()-1;
			if(r[x[i]]>r[y[i]]) {
				penal[x[i]]++;
			}
			if(r[x[i]]<r[y[i]]) {
				penal[y[i]]++;
			}
		}
		TreeMap<Integer, Integer> map = new TreeMap<>(); 
		for(int i=0; i<n; i++) {
			if(map.containsKey(r[i])) {
				map.put(r[i], map.get(r[i])+1);
			} else {
				map.put(r[i], 1);
			}
		}
		HashMap<Integer, Integer> tbl = new HashMap<>();
		{
			int acc = 0;
			for(Map.Entry<Integer, Integer> e : map.entrySet()) {
				tbl.put(e.getKey(), acc);
				acc+= e.getValue();
			}	
		}
		System.out.print(tbl.get(r[0])-penal[0]);
		for(int i=1; i<n; i++) {
			System.out.print(" " + (tbl.get(r[i])-penal[i]));
		}
		System.out.println();
	}

	static class MyScanner
	{
		BufferedReader br;
		StringTokenizer st;
		public MyScanner(InputStream s)
		{
			br=new BufferedReader(new InputStreamReader(s));
		}
		public String nextLine() throws IOException
		{
			return br.readLine();
		}
		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 double nextDouble() throws IOException
		{
			return Double.parseDouble(next());
		}
		public boolean ready() throws IOException
		{
			return br.ready();
		}
		public long nextLong() throws IOException
		{
			return Long.parseLong(next());
		}
	}
}