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.InputStreamReader;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {

        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] temp = in.readLine().split(" ");

        int n = Integer.parseInt(temp[0]);
        int k = Integer.parseInt(temp[1]);

        TreeMap<Integer, ArrayList<Integer>> m = new TreeMap<>();
        Integer[] r = new Integer[n];
        temp = in.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            r[i] = Integer.parseInt(temp[i]);
            if (m.containsKey(r[i])) {
                m.get(r[i]).add(i);
            } else {
                m.put(r[i], new ArrayList<>());
                m.get(r[i]).add(i);
            }
        }


        int[] remove = new int[n];

        for (int i = 0; i < k; i++) {
            temp = in.readLine().split(" ");
            int a = Integer.parseInt(temp[0]) - 1;
            int b = Integer.parseInt(temp[1]) - 1;
            if (r[a] > r[b]) remove[a]++;
            if (r[b] > r[a]) remove[b]++;
        }
        Arrays.sort(r);

        int prev = -1;
        int p = 0;
        for (int i: r) {
            if (i == prev) {
                prev = i;
                p++;
                continue;
            }

            for (int ind: m.get(i)) {
                remove[ind] = p - remove[ind];
            }

            prev = i;
            p++;
        }

        for (int i: remove) {
            System.out.printf(i + " ");
        }
        System.out.println();

    }

}