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.math.*;
import java.util.*;
import java.io.*;

public class Test5 {

    static int[] m, n, st, o;

    public static void main(String[] z) throws Exception{
        Scanner s = new Scanner(System.in);
        int a = s.nextInt(), b = s.nextInt(),x,y;
        m = new int[a];
        st = new int[a];
        n = new int[a];
        o = new int[a];
        for(int q=0; q<a; q++){
            m[q] = s.nextInt();
            st[q] = m[q];
            n[q] = q;
        }
        PyraSort.sort(m);
        for(int q=0; q<a; q++){
            if(q==0) o[n[q]]=0;
            else{
                if(m[q]==m[q-1]) o[n[q]] = o[n[q-1]];
                else o[n[q]]=q;
            }
        }
        for(int q=0; q<b; q++){
            x = s.nextInt();
            y = s.nextInt();
            if(st[x-1]>st[y-1]) o[x-1]--;
            else if(st[y-1]>st[x-1])o[y-1]--;
        }
        for(int q=0; q<a; q++) System.out.print(o[q]+" ");
    }

    static class PyraSort {

        private static int heapSize;

        public static void sort(int[] a) {
            buildHeap(a);
            while (heapSize > 1) {
                swap(a, 0, heapSize - 1);
                heapSize--;
                heapify(a, 0);
            }
        }

        private static void buildHeap(int[] a) {
            heapSize = a.length;
            for (int i = a.length / 2; i >= 0; i--) {
                heapify(a, i);
            }
        }

        private static void heapify(int[] a, int i) {
            int l = 2 * i + 2;
            int r = 2 * i + 1;
            int largest = i;
            if (l < heapSize && a[i] < a[l]) {
                largest = l;
            }
            if (r < heapSize && a[largest] < a[r]) {
                largest = r;
            }
            if (i != largest) {
                swap(a, i, largest);
                heapify(a, largest);
            }
        }

        private static void swap(int[] a, int i, int j) {
            a[i] ^= a[j] ^= a[i];
            a[j] ^= a[i];
            n[i] ^= n[j] ^= n[i];
            n[j] ^= n[i];
        }
    }
}
