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.lang.reflect.Array;
import java.util.*;
import java.io.*;
/*
11 20
4 3 7 -3 -14 1 6 8 -3 -9 -6
*/
public class Main {
    static class Node{
        int index;
        long value;
        Node(int index,long value){
            this.index=index;
            this.value=value;
        }
    }

    public static void main(String[] args) throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int k=Integer.parseInt(st.nextToken());
        Node[] arr=new Node[n];
        st=new StringTokenizer(br.readLine());
        long[] a=new long[n];
        for(int i=0;i<n;i++){
            arr[i]=new Node(i,Long.parseLong(st.nextToken()));
            a[i]=arr[i].value;
        }
        Arrays.sort(arr, new Comparator<Node>() {
            @Override
            public int compare(Node o1, Node o2) {
                if(o1.value==o2.value){
                    return o1.index-o2.index;
                }else {
                    return Long.compare(o1.value,o2.value);
                }
            }
        });
        int[] count=new int[n];
        for(int i=0;i<n;i++){
            long val=arr[i].value;
            int temp=i;
            while (i<n&&val==arr[i].value){
                count[arr[i].index]=temp;
                i++;
            }
            i--;
        }
        for(int i=0;i<k;i++){
            st=new StringTokenizer(br.readLine());
            int l=Integer.parseInt(st.nextToken())-1;
            int r=Integer.parseInt(st.nextToken())-1;
            if(a[l]>a[r]){
                count[l]--;
            }else if(a[r]>a[l]){
                count[r]--;
            }
        }
        StringBuilder str=new StringBuilder("");
        for(int i=0;i<n;i++){
            str.append(count[i]+" ");
        }
        pw.println(str);


        pw.close();
    }
}