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.util.Arrays;
import java.util.Scanner;

public class F {
    private static int[] mergesort(int[] array1) {
        int[] buffer1 = Arrays.copyOf(array1, array1.length);
        int[] buffer2 = new int[array1.length];
        return mergesortInner(buffer1, buffer2, 0, array1.length);
    }

    private static int[] mergesortInner(int[] buffer1, int[] buffer2,
                                        int startIndex, int endIndex) {
        if (startIndex >= endIndex - 1) {
            return buffer1;
        }
        int middle = startIndex + (endIndex - startIndex) / 2;
        int[] sorted1 = mergesortInner(buffer1, buffer2, startIndex, middle);
        int[] sorted2 = mergesortInner(buffer1, buffer2, middle, endIndex);
        int index1 = startIndex;
        int index2 = middle;
        int destIndex = startIndex;
        int[] result = sorted1 == buffer1 ? buffer2 : buffer1;
        while (index1 < middle && index2 < endIndex) {
            result[destIndex++] = sorted1[index1] < sorted2[index2]
                    ? sorted1[index1++] : sorted2[index2++];
        }
        while (index1 < middle) {
            result[destIndex++] = sorted1[index1++];
        }
        while (index2 < endIndex) {
            result[destIndex++] = sorted2[index2++];
        }
        return result;
    }

    private static int binarySerach(int[] a, int lo, int hi, int key) {
        while (hi - lo > 1) {
            int m = lo + (hi - lo) / 2;
            if (a[m] >= key) {
                hi = m;
            } else {
                lo = m;
            }
        }
        return hi;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int[] r = new int[n];
        for (int i = 0; i < n; i++) {
            r[i] = in.nextInt();
        }
        mergesort(r);
        int[] arg = new int[n];
        for (int i = 0; i < k; i++) {
            int x = in.nextInt() - 1;
            int y = in.nextInt() - 1;
            if (r[x] > r[y]) arg[x]++;
            else if (r[y] > r[x]) arg[y]++;
        }
        int[] rSorted = mergesort(Arrays.copyOf(r, n));
        for (int i = 0; i < n; i++) {
            System.out.print(binarySerach(rSorted, -1, n, r[i]) - arg[i] + " ");
        }
    }
}