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.DataInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 public class CF_triangle{
    static long MOD = 1000000007;
    static int modulo=998244353;
    static int parent[]=new int[26];
    static int rank[]=new int[26];
    static boolean visited[];
    static Scanner in;
    public static void main(String[] args) {
       Scanner in=new Scanner(System.in);
       int n=in.nextInt();
       int k=in.nextInt();
       ArrayList<Integer> list=new ArrayList<>();
        ArrayList<Integer> list1=new ArrayList<>();
       for(int i=0;i<n;i++){
           int a=in.nextInt();
           list.add(a);
           list1.add(a);
       }
              
       int count[]=new int[n];
       while(k-->0){
           int x=in.nextInt()-1;
           int y=in.nextInt()-1;
           if(list.get(x)>list.get(y))
               count[x]++;
           if(list.get(y)>list.get(x))
               count[y]++;
               
       }
       Collections.sort(list);
       for(int i=0;i<n;i++){
           int pos=bs(list1.get(i)-1,list)+1;
           System.out.print(pos-count[i]+" ");
       }
       }
    public static int bs(int val,ArrayList<Integer> list){
        int l=0;
        int r=list.size()-1;
        int ans=-1;
        while(l<=r){
            int m=(l+r)/2;
            if(list.get(m)<=val){
                ans=m;
                l=m+1;
            }
            else{
                r=m-1;
            }
        }
        return ans;
    }
 }