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 Main {
    public static int binarysearch(ArrayList<Integer> al1,int value) {
        int low=0;
        int high=al1.size()-1;
        int res=-1;
        while(low<=high) {
            int mid=low+(high-low)/2;
            if(al1.get(mid)<=value) {
                res=mid;
                low=mid+1;
            }
            else 
            high=mid-1;
        }
        return res;
    }
    
    public static void main(String[] args) throws IOException {
      //  try {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        ArrayList<Integer> al1=new ArrayList<>();
        ArrayList<Integer> al2=new ArrayList<>();
        for(int i=0;i<n;i++) {
            int ele=sc.nextInt();
            al1.add(ele);
            al2.add(ele);
        }
        int notvis[]=new int[n];
        for(int i=1;i<=k;i++) {
            int x=sc.nextInt();
            int y=sc.nextInt();
            x=x-1;
            y=y-1;
            if(al1.get(x)>al1.get(y)) 
            notvis[x]++;
             if(al1.get(y)>al1.get(x)) 
            notvis[y]++;
        
        }
        Collections.sort(al1);
        for(int i=0;i<n;i++) {
            int numberlessthan=binarysearch(al1,al2.get(i)-1)+1;
            System.out.print(numberlessthan-notvis[i]+" ");
        }
        System.out.println();
  //  }catch(Exception e) {}
    }
}
        