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.
//package com.company;
import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
    public static Long [] r;
    public static Long [] v;
    public static int n;
    public static ArrayList<Long>[] adj;
    
    public static void main(String[] args) throws IOException {
        InputReader sc = new InputReader(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        run(1, sc, pw);
        pw.close();
    }
    public static void run(int testCases, InputReader sc, PrintWriter pw){

        for (int tst = 1; tst <= testCases; tst++){
          
          n = sc.nextInt();
          int k = sc.nextInt();
          adj = new ArrayList[n];
          for(int i = 0; i < n; i++)
            adj[i] = new ArrayList<Long>();
          r = new Long[n];
          v = new Long[n];
          for(int i = 0; i < n; i++){
            r[i] = sc.nextLong();
            v[i] = r[i];
          }            
          while(k-- > 0){
            
            int x, y;
            
            x = sc.nextInt() - 1;
            y = sc.nextInt() - 1;
            
            adj[x].add(r[y]);
            adj[y].add(r[x]);
            
          }
          
          Arrays.sort(r);
          
          for(int i = 0; i < n; i++){
            
	         Collections.sort(adj[i]);
            int tot = 0;
            int lo = 0, hi = n-1;
            
            while(lo < hi){
              
              int md = lo + (hi - lo + 1) / 2;
              
              if(r[md] < v[i])lo = md;
              else
                hi = md - 1;
              
            }
            
            if(r[lo] < v[i])tot =lo+1;
            if(adj[i].size() > 0){
            lo = 0;
            hi = adj[i].size() - 1;
            
            while(lo < hi){
              
              int md = lo + (hi-lo+1) / 2;
              
              if(adj[i].get(md) < v[i])lo=md;
              else hi = md - 1;
              
            }
            if(adj[i].get(lo) < v[i])tot -= (lo+1);
            }pw.print(tot + " ");
          }
          
        }
    }
    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer stt;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public String nextString() {
            while (stt == null || !stt.hasMoreTokens()) {
                stt = new StringTokenizer(nextLine());
            }
            return stt.nextToken();
        }

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

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

    }
}