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.IOException;
import java.io.InputStreamReader;
import java.util.List;
import java.util.TreeMap;
import java.util.StringTokenizer;
import java.util.Map;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStream;

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

    static class F {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.ni(), k = in.ni();
            int[] a = in.na(n);
            int[][] g = in.readGraph(n, k);
            TreeMap<Integer, List<Integer>> map = new TreeMap<>();
            for (int i = 0; i < n; i++) {
                List<Integer> list = map.computeIfAbsent(a[i], k1 -> new LinkedList<>());
                list.add(i);
            }
            int[] cnts = new int[n];
            int num = 0;
            int[] ans = new int[n];
            for (Integer r : map.keySet()) {
                List<Integer> list = map.get(r);
                if (num > 0) {
                    for (Integer id : list) {
                        if (num - cnts[id] > 0)
                            ans[id] = num - cnts[id];
                    }
                }

                num += list.size();
                for (Integer id : list) {
                    for (int v : g[id]) {
                        cnts[v]++;
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                out.print(ans[i] + " ");
            }
            out.println();
        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String ns() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(ns());
        }

        public int[] na(int n) {
            int[] ret = new int[n];
            for (int i = 0; i < n; i++) {
                ret[i] = ni();
            }
            return ret;
        }

        public int[][] readGraph(int n, int m) {
            int[][] g = new int[n][];
            int[] from = new int[m];
            int[] to = new int[m];
            int[] sizes = new int[n];
            for (int i = 0; i < m; i++) {
                int u = ni() - 1;
                int v = ni() - 1;
                from[i] = u;
                to[i] = v;
                sizes[u]++;
                sizes[v]++;
            }
            for (int i = 0; i < n; i++) {
                g[i] = new int[sizes[i]];
            }
            for (int i = 0; i < m; i++) {
                g[from[i]][--sizes[from[i]]] = to[i];
                g[to[i]][--sizes[to[i]]] = from[i];
            }
            return g;
        }

    }
}

