F. Mentors

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 $k$ $(2\le n\le 2\cdot {10}^5$, $0\le k\le min(2\cdot {10}^5,\frac{n\cdot (n-1)}{2}))$

— total number of programmers and number of pairs of programmers which are in a quarrel.

The second line contains a sequence of integers $r_1,r_2,\dots ,r_n$

$(1\le r_i\le {10}^9)$, where $r_i$ equals to the skill of the $i$

-th programmer.

Each of the following $k$

lines contains two distinct integers $x$, $y$ $(1\le x,y\le n$, $x\ne y)$ — 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

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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn=2e5+10;

int ans[maxn];

struct node{
    int id,val;
    node(){}
    node(int _val){
        val=_val;
    }
    bool operator<(const node& b)const{
        return val<b.val;
    }
}p[maxn];
/*
暴力统计一下数量就好了
*/

int main()
{
   int n,k;
   scanf("%d %d",&n,&k);
   for(int i=1;i<=n;i++){
    scanf("%d",&p[i].val);
     p[i].id=i;
   }

   for(int i=1;i<=k;i++){
      int u,v;
      scanf("%d %d",&u,&v);
      if(p[u].val>p[v].val)
        ans[u]--;
      if(p[v].val>p[u].val)
        ans[v]--;
   }
   sort(p+1,p+n+1);
   for(int i=1;i<=n;i++){
     int pos=lower_bound(p+1,p+n+1,node(p[i].val))-p;
     ans[p[i].id]+=pos-1;
   }
   for(int i=1;i<=n;i++){
    if(i==1)printf("%d",ans[i]);
    else printf(" %d",ans[i]);
   }
   printf("\n");
   return 0;
}