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 N=1e5+50,mod=1e9+7;
int n,k,m,a[N],c[N],b[N],J[N],I[N],ans;
int C(int n,int m){return n<m?0:1ll*J[n]*I[m]%mod*I[n-m]%mod;}
int main(){
    // freopen("test.in","r",stdin);
    scanf("%d%d",&n,&k);J[0]=I[0]=I[1]=1;
    for(int i=2;i<=n;i++)I[i]=mod-1ll*mod/i*I[mod%i]%mod;
    for(int i=1;i<=n;i++)J[i]=1ll*J[i-1]*i%mod,I[i]=1ll*I[i-1]*I[i]%mod;
    for(int i=1;i<=n;i++)scanf("%d",&a[i]),c[i]=a[i];
    sort(c+1,c+n+1);m=unique(c+1,c+n+1)-c-1;
    for(int i=1;i<=n;i++){
        a[i]=lower_bound(c+1,c+m+1,a[i])-c;
        b[a[i]]++;
    }
    for(int i=1,sum=0;i<=m;i++){
        ans=(ans+1ll*(C(sum+b[i],k)-C(sum,k))*c[i])%mod;
        sum+=b[i];
    }
    for(int i=m,sum=0;i;i--){
        ans=(ans-1ll*(C(sum+b[i],k)-C(sum,k))*c[i])%mod;
        sum+=b[i];
    }
    printf("%d\n",(ans+mod)%mod);
    return 0;
}