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
/* 
* ID: juryc
* PROG: MIN-MAX-SUMS
* LANG: C++
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
int a[N]; 
ll fact[N];
ll invfact[N];
ll mod=1e9+7;
ll modpow(ll a,ll b){
ll res=1; a%=mod;
while(b){
if(b&1) { res=(res*a)%mod; }
a=(a*a)%mod;
b>>=1;
}
return res;
}

ll inv(ll a){
return modpow(a,mod-2);
}

void compute(){
fact[0]=invfact[0]=1;
for(int i=1;i<N;i++){
fact[i]=(fact[i-1]*i)%mod;
invfact[i]=inv(fact[i]);
}
}

ll choose(ll n,ll r){
if(n<0||r<0||n<r) return 0;
if(n==r) return 1;
if(r==0)return 1;
return ((fact[n]*invfact[r])%mod*invfact[n-r])%mod;
}

int main(){
int n,k; cin>>n>>k;
for(int i=1;i<=n;i++){ cin>>a[i]; }
sort(a+1,a+1+n);
compute();
ll ans=0;
for(int i=1;i<=n;i++){
ans=(ans+(a[i]*choose(i-1,k-1))%mod)%mod;
ans=(ans-(a[i]*choose(n-i,k-1))%mod+mod)%mod;
}
cout<<ans<<"\n";
return 0;
}


