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 codeforces{
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

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

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    static class p implements Comparable{
        long sk;long id;
        p(long sk,long id){
            this.sk=sk;this.id=id;
        }
        @Override
        public int compareTo(Object o) {
            long s = ((p)o).sk;
            return (int)(this.sk-s);
            //return 0;
        }
    }
    public static void main(String[] args){
        FastReader in = new FastReader();OutputStream outputStream = System.out;PrintWriter out = new PrintWriter(outputStream);
        long n = in.nextLong();long k = in.nextLong();p[] arr = new p[(int)n];
        int q=0;
        for (int i=0;i<n;i++) {
            arr[i] = new p(in.nextLong(),i);
            if (i!=0&&q==0&&arr[i-1].sk!=arr[i].sk)q=1;
        }
        if (q==0) {for (int i=0;i<n;i++) out.print(0+" ");out.close();return;}
        long[] pr = new long[(int)n];
        for (int i=0;i<k;i++){
            long x = in.nextLong()-1;long y = in.nextLong()-1;
            if (arr[(int)x].sk>arr[(int)y].sk) pr[(int)x]--;
            else if (arr[(int)x].sk<arr[(int)y].sk) pr[(int)y]--;
        }
        Arrays.sort(arr);//long ans=0;
        //out.print(0+" ");
        long mp=0;
        for (int i=1;i<arr.length;i++){
            if (arr[i-1].sk==arr[i].sk) {
                mp++;pr[(int) arr[i].id] -= mp;
            }
            else mp=0;
        }
        for (int i=1;i<n;i++){
            long ans=0;
            ans+=i;//int j=i-1;
            //while (j>=0&&arr[j].sk==arr[i].sk) {ans--;j--;}
            pr[(int)arr[i].id]+=ans;
            //out.print(ans+" ");]
        }
        for (int i=0;i<pr.length;i++) out.print(pr[i]+" ");
        out.close();
    }
}