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F. Mentors

 

In BerSoft nn programmers work, the programmer ii is characterized by a skill riri.

A programmer aa can be a mentor of a programmer bb if and only if the skill of the programmer aa is strictly greater than the skill of the programmer bb (ra>rb)(ra>rb) and programmers aa and bb are not in a quarrel.

You are given the skills of each programmers and a list of kk pairs of the programmers, which are in a quarrel (pairs are unordered). For each programmer ii, find the number of programmers, for which the programmer ii can be a mentor.

Input

The first line contains two integers nn and kk (2≤n≤2⋅105(2≤n≤2⋅105, 0≤k≤min(2⋅105,n⋅(n−1)2))0≤k≤min(2⋅105,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 r1,r2,…,rnr1,r2,…,rn (1≤ri≤109)(1≤ri≤109), where riri equals to the skill of the ii-th programmer.

Each of the following kk lines contains two distinct integers xx, yy (1≤x,y≤n(1≤x,y≤n, x≠y)x≠y) — pair of programmers in a quarrel. The pairs are unordered, it means that if xx is in a quarrel with yy then yy is in a quarrel with xx. Guaranteed, that for each pair (x,y)(x,y) there are no other pairs (x,y)(x,y) and (y,x)(y,x) in the input.

Output

Print nn integers, the ii-th number should be equal to the number of programmers, for which the ii-th programmer can be a mentor. Programmers are numbered in the same order that their skills are given in the input.

Examples

input

Copy

4 2
10 4 10 15
1 2
4 3

output

Copy

0 0 1 2 

input

Copy

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5

output

Copy

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.

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
using namespace std;

int ans[200005];
int a[200005];
int b[200005];
int N[200005];

int cmp(int a,int b){
	return N[a]<N[b];
}

int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		scanf("%d",&N[i]);
		a[i] = i;
	}
	for(int i=0;i<m;i++){
		int x,y;
		cin>>x>>y;
		if(N[x]>N[y]){
			ans[x]++;
		}
		else if(N[x]<N[y]){
			ans[y]++;
		}
	} 
	stable_sort(a+1,a+n+1,cmp);
	for(int i=2;i<=n;i++){
		if(N[a[i]]>N[a[i-1]]){
			b[a[i]] = i-1;
		}
		else{
			b[a[i]] = b[a[i-1]];
		}
	}
	for(int i=1;i<=n;i++){
    	if(i==1)
			printf("%d",b[i]-ans[i]);
    	else 
			printf(" %d",b[i]-ans[i]);
   }
	return 0;
}