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.
// Don't place your source in a package
import javax.swing.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        //BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

        //reading /writing file
        //Scanner sc=new Scanner(new File("input.txt"));
        //PrintWriter pr=new PrintWriter("output.txt");


        int T=1;
        for(int t=0;t<T;t++){
            int n=Int();int k=Int();
            int A[]=Arr(n);
            int B[][]=new int[k][2];
            for(int i=0;i<k;i++){
                B[i][0]=Int()-1;
                B[i][1]=Int()-1;
            }

            Solution sol=new Solution(out);
            sol.solution(A,B);
        }
        out.close();

    }

    public static int[] Arr(int n){
        int A[]=new int[n];
        for(int i=0;i<n;i++)A[i]=Int();
        return A;
    }
    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}








class Solution{
    PrintWriter out;
    int INF=Integer.MAX_VALUE;
    int MOD=1000000007;
    public Solution(PrintWriter out){
        this.out=out;
    }



    public void solution(int A[],int B[][]){
        List<Integer>g[]=new ArrayList[A.length];
        int C[]=new int[A.length];
        for(int i=0;i<A.length;i++)C[i]=A[i];
        sort(C);
        Arrays.setAll(g,e->new ArrayList<>());

        for(int p[]:B){
            int u=p[0],v=p[1];
            g[u].add(v);
            g[v].add(u);
        }


        for(int i=0;i<A.length;i++){
            List<Integer>list=g[i];
            int cnt=0;
            for(int index:list){
                if(A[index]<A[i])cnt++;
            }

            int l=0,r=A.length-1;
            int pos=-1;
            while(l<=r){
                int mid=l+(r-l)/2;
                if(C[mid]<A[i]){
                    pos=mid;
                    l=mid+1;
                }
                else{
                    r=mid-1;
                }
            }

            if(pos!=-1){
                out.println(((pos+1)-cnt)+" ");
            }
            else{
                out.print(0+" ");
            }

        }

    }

    public void sort(int A[]){
        PriorityQueue<Integer>pq=new PriorityQueue<>();
        for(int i:A)pq.add(i);
        for(int i=0;i<A.length;i++)A[i]=pq.poll();
    }


}



