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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.StringTokenizer;
import javafx.util.Pair;

public class Solve4 {

    public static void main(String[] args) throws IOException {
        new Solve4().solve();
    }

    public void solve() throws IOException {
        FastReader sc = new FastReader();
        PrintWriter pw = new PrintWriter(System.out);
        int n = sc.nextInt(), k = sc.nextInt();
        ArrayList<Pair<Integer, Integer>> a = new ArrayList();
        for (int i = 0; i < n; i++) {
            a.add(new Pair(i, sc.nextInt()));
        }
        int[] q = new int[n];
        for (int i = 0; i < k; i++) {
            int x = sc.nextInt(), y = sc.nextInt();
            if (a.get(x - 1).getValue() > a.get(y - 1).getValue()) {
                q[x - 1]++;
            } else if (a.get(y - 1).getValue() > a.get(x - 1).getValue()) {
                q[y - 1]++;
            }
        }
        Comparator<Pair<Integer, Integer>> cmp = new Comparator<Pair<Integer, Integer>>() {
            @Override
            public int compare(Pair<Integer, Integer> o1, Pair<Integer, Integer> o2) {
                return o1.getValue().compareTo(o2.getValue());
            }
        };
        Collections.sort(a, cmp);
        int s = 1, temp = 0, ans[] = new int[n];
        for (int i = 1; i < n; i++, s++) {
            if (a.get(i).getValue() > a.get(i - 1).getValue()) {
                temp = s;
            }
            if (temp != 0) {
                ans[a.get(i).getKey()] = temp - q[a.get(i).getKey()];
            }
        }
        pw.write(ans[0] + "");
        for (int i = 1; i < n; i++) {
            pw.write(" " + ans[i]);
        }
        pw.println();
        pw.flush();
        pw.close();
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

        public String next() throws IOException {
            if (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

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

        public Long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public Double nextouble() throws IOException {
            return Double.parseDouble(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }
    }
}
