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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;

const int N=100005,P=1000000007;
int n,k,a[N],Ans;
int Fac[N],Inv[N];

ll Pow(ll a,int b,ll s=1){for(;b;b>>=1,a=a*a%P)if(b&1)s=s*a%P;return s%P;}
inline ll C(int n,int m){return (ll)Fac[n]*Inv[m]%P*Inv[n-m]%P;}

int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;++i)scanf("%d",&a[i]);
    for(int i=Fac[0]=1;i<=n;++i)Fac[i]=(ll)Fac[i-1]*i%P;
    Inv[n]=Pow(Fac[n],P-2);
    for(int i=n;i>=1;--i)Inv[i-1]=(ll)Inv[i]*i%P;
    std::sort(a+1,a+n+1);
    for(int i=k;i<=n;++i)Ans=(Ans+(ll)a[i]*C(i-1,k-1))%P;
    for(int i=1;i<=n-k+1;++i)Ans=(Ans-(ll)a[i]*C(n-i,k-1))%P;
    printf("%d\n",(Ans+P)%P);
    return 0;
}
