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.*;
import java.util.*;
public class F
{
    private static class Person
    {
        private int pos,ans,val;

        Person(int val, int pos)
        {
            this.val=val;
            this.pos=pos;
            ans=0;
        }
    }

    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N,K,x,y,count;

        String s[]=br.readLine().trim().split(" ");

        N=Integer.parseInt(s[0]);
        K=Integer.parseInt(s[1]);

        s=br.readLine().trim().split(" ");

        Person a[]=new Person[N];
        for(i=0;i<N;i++)
          a[i]=new Person(Integer.parseInt(s[i]),i);

        while(K-->0)
        {
            s=br.readLine().trim().split(" ");

            x=Integer.parseInt(s[0])-1;
            y=Integer.parseInt(s[1])-1;

            if(a[x].val<a[y].val)
                a[y].ans--;
            else if(a[x].val>a[y].val)
                a[x].ans--;
        }

        Arrays.sort(a,(Person u,Person v)-> v.val-u.val);

        count=0;
        for(i=N-2;i>=0;i--)
        {
            if(a[i+1].val==a[i].val)
            {
                count++;
                a[i].ans-=count;
            }
            else
                count=0;
        }

        for(i=0;i<N;i++)
        {
            a[i].ans+=N-1-i;
        }

        Arrays.sort(a, (Person u, Person v)-> u.pos-v.pos);

        StringBuilder sb=new StringBuilder();
        for(i=0;i<N;i++)
            sb.append(a[i].ans).append(" ");

        System.out.println(sb);
    }
}
