For a finite set of integers X, let f(X)=\max X - \min X.

Given are N integers A_1,...,A_N.

We will choose K of them and let S be the set of the integers chosen. If we distinguish elements with different indices even when their values are the same, there are {}_N C_K ways to make this choice. Find the sum of f(S) over all those ways.

Since the answer can be enormous, print it \bmod (10^9+7).

Constraints

* 1 \leq N \leq 10^5
* 1 \leq K \leq N
* |A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:


N K
A_1 ... A_N


Output

Print the answer \bmod (10^9+7).

Examples

Input

4 2
1 1 3 4


Output

11


Input

6 3
10 10 10 -10 -10 -10


Output

360


Input

3 1
1 1 1


Output

0


Input

10 6
1000000000 1000000000 1000000000 1000000000 1000000000 0 0 0 0 0


Output

999998537
#include<cstdio>
#include<bits/stdc++.h>
#include<algorithm>
using namespace std;
#define N 100001
const int mod=1e9+7;
int a[N],inv[N];
int Pow(int p,int q)
{
    int s=1;
    for(;q;q>>=1,p=1LL*p*p%mod)
        if(q&1) s=1LL*s*p%mod;
    return s;
}
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;++i) cin>>a[i];
    sort(a+1,a+n+1);
    for(int i=1;i<=n;++i) inv[i]=Pow(i,mod-2); 
    long long p=1,ans=a[k];
    for(int i=k+1;i<=n;++i)
    {
         p=p*(i-1)%mod;
         p=p*inv[i-k]%mod;
         ans=(ans+a[i]*p%mod)%mod;
    }
    ans-=a[n-k+1];
    p=1;
    for(int i=k+1;i<=n;++i)
    {
         p=p*(i-1)%mod;
         p=p*inv[i-k]%mod;
         ans=(ans-a[n-i+1]*p%mod)%mod;
    }
    if(ans<0) ans+=mod;
    cout<<ans;
}