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.Map;
import java.util.Scanner;
import java.util.HashMap;
import java.util.LinkedList;

/**
 * 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;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        FMentors solver = new FMentors();
        solver.solve(1, in, out);
        out.close();
    }

    static class FMentors {
        public void solve(int testNumber, Scanner in, PrintWriter out) {

            int n = in.nextInt();
            int k = in.nextInt();
            in.nextLine();

            int[] r = new int[n];

            for (int i = 0; i < n; i++)
                r[i] = in.nextInt();
            in.nextLine();

            LinkedList<Integer>[] quarrels = new LinkedList[n];

            for (int i = 0; i < n; i++) {
                quarrels[i] = new LinkedList<Integer>();
            }

            for (int i = 0; i < k; i++) {
                int left = in.nextInt() - 1;
                int right = in.nextInt() - 1;


                quarrels[left].add(right);
                quarrels[right].add(left);
                in.nextLine();
            }


            Map<Integer, Integer> map = new HashMap<>();

            Integer[] rCopy = new Integer[n];

            for (int i = 0; i < n; i++)
                rCopy[i] = new Integer(r[i]);

            Arrays.sort(rCopy);

            { //populate map
                int key = rCopy[0];
                int count = 0;
                int previousKeyCount = 0;
                map.put(key, count);
                for (int i = 1; i < n; i++) {
                    key = rCopy[i];
                    count = i;
                    int previousKey = rCopy[i - 1];
                    if (previousKey == key) {
                        previousKeyCount++;
                        map.put(key, count - previousKeyCount);
                    } else {
                        previousKeyCount = 0;
                        map.put(key, count);
                    }
                }
            }

            for (int i = 0; i < n; i++) {
                int rank = r[i];
                int ranksLessThanCount = map.get(rank);

                int count = 0;
                for (Integer programmerInQuarrelWith : quarrels[i]) {
                    //find programmer in quarrel with that have a
                    //rank less than programmer i's rank
                    int pInQuarrelWithRank = r[programmerInQuarrelWith];

                    if (pInQuarrelWithRank < rank)
                        count++;
                }

                int ranksLessThanCountNotIncludingQuarrels = ranksLessThanCount - count;
                int partialSol = ranksLessThanCountNotIncludingQuarrels;

                out.print(partialSol + " ");
            }
        }

    }
}

