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;
#define rep(i,a,b) for(int i=int(a);i<int(b);i++)
typedef long long ll;

const int MAX = 100010;
const ll mod = 1e9+7;

ll fac[MAX], finv[MAX], inv[MAX];

void init() {
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[0]=inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%mod;
        inv[i]=mod-inv[mod%i]*(mod/i)%mod;
        finv[i]=finv[i-1]*inv[i]%mod;
    }
}


int main(){
    ll n,k;cin>>n>>k;
    if(k==0){
        cout<<0<<endl;
        return 0;
    }
    ll a[100010];
    rep(i,0,n)cin>>a[i];
    sort(a,a+n);
    ll an=0;
    init();
    rep(i,k-1,n){
        an+=(((a[i]*fac[i])%mod*finv[k-1])%mod*finv[i-k+1])%mod;
    }
    rep(i,0,n-k+1){
        an-=(((a[i]*fac[n-i-1])%mod*finv[k-1])%mod*finv[n-i-k])%mod;
    }
    an=(an+mod)%mod;
    cout<<an<<endl;
}
    







