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
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
int a[100007];
const long long mod = 1e9+7;
long long inv[100007];
long long qpow(int p,int q){
    long long s=1;
    for(;q;q>>=1,p=1ll*p*p%mod)
        if(q&1)
            s=1ll*s*p%mod;
    return s;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;++i)
        cin>>a[i];
    sort(a+1,a+1+n);
    for(int i=1;i<=n;++i)
        inv[i]=qpow(i,mod-2);
    long long sum=0;
    long long temp=1;
    for(int i=k;i<=n;++i){
        sum=(sum+a[i]*temp%mod)%mod;
        temp=temp*i%mod;
        temp=temp*inv[i+1-k]%mod;
    }
    temp=1;
    for(int i=n-k+1;i;--i){
        sum=(sum+mod-a[i]*temp%mod)%mod;
        temp=temp*(n-i+1)%mod;
        temp=temp*inv[n-i+2-k]%mod;
    }
    cout<<sum;
    return 0;
}
