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.util.Arrays;
import java.util.StringTokenizer;
import java.util.Random;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

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

    static class TaskF {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] r = new int[n];
            int[] ar = new int[n];
            for (int i = 0; i < n; i++) {
                r[i] = in.nextInt();
                ar[i] = r[i];
            }
            shuffleArray(ar);
            Arrays.sort(ar);

            int[] ans = new int[n];

            for (int i = 0; i < n; i++) {
                int p = binSearch(ar, r[i]);
                if (p != -1)
                    ans[i] = p + 1;
            }

            for (int i = 0; i < k; i++) {
                int p1 = in.nextInt() - 1;
                int p2 = in.nextInt() - 1;
                if (r[p2] > r[p1])
                    ans[p2]--;
                else if (r[p1] > r[p2])
                    ans[p1]--;
            }

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

        private static void shuffleArray(int[] array) {
            int index;
            Random random = new Random();
            for (int i = array.length - 1; i > 0; i--) {
                index = random.nextInt(i + 1);
                if (index != i) {
                    array[index] ^= array[i];
                    array[i] ^= array[index];
                    array[index] ^= array[i];
                }
            }
        }

        int binSearch(int[] ar, int a) {
            int l = -1;
            int r = ar.length;
            while (r - l > 1) {
                int m = (l + r) / 2;
                if (ar[m] < a)
                    l = m;
                else
                    r = m;
            }
            return l;
        }

    }

    static class InputReader {
        StringTokenizer st;
        BufferedReader br;

        public InputReader(InputStream is) {
            BufferedReader br = new BufferedReader(new InputStreamReader(is));
            this.br = br;
        }

        public String next() {
            if (st == null || !st.hasMoreTokens()) {
                String nextLine = null;
                try {
                    nextLine = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (nextLine == null)
                    return null;
                st = new StringTokenizer(nextLine);
            }
            return st.nextToken();
        }

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

    }
}

