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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.BufferedOutputStream;
import java.util.stream.Stream;
import java.io.Writer;
import java.io.BufferedReader;
import java.util.regex.Pattern;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author amalev
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        FastWriter out = new FastWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public void solve(int testNumber, FastScanner in, FastWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] r = in.nextIntArray();
            TaskE.Pair[] set = new TaskE.Pair[n];
            for (int i = 0; i < n; i++) {
                set[i] = new TaskE.Pair(r[i], i);
            }
            Arrays.sort(set,
                    (x, y) -> {
                        int res = Integer.compare(x.r, y.r);
                        if (res == 0) {
                            res = Integer.compare(x.i, y.i);
                        }
                        return res;
                    });

            int[] dec = new int[n];
            for (int i = 0; i < k; i++) {
                int x = in.nextInt() - 1;
                int y = in.nextInt() - 1;
                if (r[x] > r[y]) {
                    dec[x]++;
                } else if (r[x] < r[y]) {
                    dec[y]++;
                }
            }

            int[] res = new int[n];
            int l = 0;
            int sum = 0;
            for (int i = 0; i < n; i++) {
                TaskE.Pair p = set[i];
                if (set[l].r < set[i].r) {
                    sum += i - l;
                    l = i;
                }

                res[p.i] = sum - dec[p.i];
            }

            for (int i = 0; i < n; i++) {
                out.print(res[i]);
                if (i < n - 1) out.print(" ");
            }
        }

        static class Pair {
            int r;
            int i;

            public Pair(int r, int i) {
                this.r = r;
                this.i = i;
            }

        }

    }

    static class FastWriter extends PrintWriter {
        static final int BUFFER_SIZE = 1 << 17;

        public FastWriter(OutputStream out) {
            super(new BufferedOutputStream(out, BUFFER_SIZE));
        }

        public FastWriter(Writer out) {
            super(out);
        }

        public void close() {
            super.close();
        }

    }

    static class FastScanner {
        static final int BUFFER_SIZE = 1 << 17;
        final BufferedReader input;
        String[] buffer;
        int pos;
        final static Pattern SEPARATOR = Pattern.compile("\\s+");

        public FastScanner(InputStream inputStream) {
            input = new BufferedReader(new InputStreamReader(inputStream), BUFFER_SIZE);
        }

        private String[] readAndSplit() {
            try {
                return SEPARATOR.split(input.readLine());
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        }

        private String readToBuffer() {
            if (buffer == null || pos >= buffer.length) {
                buffer = readAndSplit();
                pos = 0;
            }
            return buffer[pos++];
        }

        public int nextInt() {
            return Integer.parseInt(readToBuffer());
        }

        public int[] nextIntArray() {
            return Arrays.stream(readAndSplit()).mapToInt(Integer::parseInt).toArray();
        }

    }
}

