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.PrintWriter;
import java.io.*;
import java.util.*;

public class TaskF {
    private static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out, 500000));
    static class MyScanner {
        /**
         * Buffered reader
         */
        private static BufferedReader br = new BufferedReader(
                new InputStreamReader(System.in));

        private static StringTokenizer st;

        /**
         * Read integer
         *
         * @return
         * @throws Exception
         */
        public static int readInt() throws Exception {
            try {
                if (st != null && st.hasMoreTokens()) {
                    return parseInt(st.nextToken());
                }
                String str = br.readLine();
                if (str != null && !str.trim().equals("")) {
                    st = new StringTokenizer(str);
                    return parseInt(st.nextToken());
                }
            } catch (IOException e) {
                close();
                return -1;
            }
            return -1;
        }

        /**
         * Read integer
         *
         * @return
         * @throws Exception
         */
        public static long readLong() throws Exception {
            try {
                if (st != null && st.hasMoreTokens()) {
                    return Long.parseLong(st.nextToken());
                }
                String str = br.readLine();
                if (str != null && !str.trim().equals("")) {
                    st = new StringTokenizer(str);
                    return Long.parseLong(st.nextToken());
                }
            } catch (IOException e) {
                close();
                return -1;
            }
            return -1;
        }

        /**
         * Read line
         * @return
         * @throws Exception
         */
        public static String readLine() throws Exception
        {
            return br.readLine();
        }

        /**
         * Parse to integer
         * @param in
         * @return integer value
         */
        public static int parseInt(String in)
        {
            // Check for a sign.
            int num  = 0, sign = -1, i = 0;
            final int len  = in.length( );
            final char ch  = in.charAt( 0 );
            if ( ch == '-' )
                sign = 1;
            else
                num = '0' - ch;

            // Build the number.
            i+=1;
            while ( i < len )
                num = num*10 + '0' - in.charAt( i++ );
            return sign * num;
        }

        /**
         * Close BufferedReader
         *
         * @throws Exception
         */
        public static void close() throws Exception {
            br.close();
        }
    }

    public static void main(String[] args) throws Exception{
        int N = MyScanner.readInt();
        int K = MyScanner.readInt();
        int[] skills = new int[N];
        int[] quarrel = new int[N];
        for(int i = 0; i < N; i ++){
            int s =  MyScanner.readInt();
            skills[i] = s;
        }
        for(int i = 0; i < K; i ++){
            int u =  MyScanner.readInt();
            int v =  MyScanner.readInt();
            if(skills[u - 1] > skills[v - 1]){
                quarrel[u - 1]++;
            } else if(skills[v - 1] > skills[u - 1]){
                quarrel[v - 1]++;
            }
        }
        //int[] temp = new int[skills.length];
        List<Integer> temp = new ArrayList<>();
        for(int s : skills){
            temp.add(s);
        }
        Collections.sort(temp);
        //System.arraycopy(skills, 0, temp, 0, skills.length);
        //Arrays.sort(temp);
        for(int i = 0; i < N; i ++){
            int s = skills[i];
            int index = binarySearch(temp, s);
            if(index > 0){
                pw.println(index - quarrel[i]);
            } else{
                pw.println(0);
            }
        }
        pw.flush();
        pw.close();
        MyScanner.close();
    }

    private static int binarySearch(List<Integer> S, int i){
        int l = 0, h = S.size() - 1;
        int ans = -1;
        while(l <= h){
            int m = l + (h - l) / 2;
            if(S.get(m) >= i){
                h = m - 1;
                ans = m;
            } else{
                l = m + 1;
            }
        }
        return ans;
    }
}
