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<bits/stdc++.h>
using namespace std;
const int mod=1000000007;
long long div(long long pow){//逆元(引数^1000000005)を返す
    long long x=1;
    for(int i=mod-2;i>0;i/=2){
        if(i%2){
            x*=pow;
            x%=mod;
        }
        pow*=pow;
        pow%=mod;
    }
    return x;
}
int main(){
    int n,k;
    cin>>n>>k;
    int a[n];
    for(int i=0;i<n;i++)cin>>a[i];
    sort(a,a+n);
    long long max=0,min=0,c=1;
    for(int i=0;i<=n-k;i++){
        max+=a[k+i-1]*c;
        min+=a[n-k-i]*c;
        max%=mod;
        min%=mod;
        c*=i+k;
        c%=mod;
        c*=div(i+1);
        c%=mod;
    }
    cout<<(max-min+mod)%mod<<endl;
    return 0;
}