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<cmath>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<cstdlib>
#define ll long long
#define lowbit(x) (x&(-x))
using namespace std;

const ll mod=1e9+7;
ll C,n,k,a[100007],ans,d[100007];

ll Ni(ll x){
    ll p=mod-2,ans=1;
    for (;p;p>>=1){
        if (p&1) ans=(ans*x)%mod;
        x=(x*x)%mod;
    }
    return ans;
}

void Init(){
    d[k-1]=1;
    for (int i=k;i<n;++i)
        d[i]=d[i-1]*i%mod*Ni(i+1-k)%mod;
}

int main(){
    scanf("%lld%lld",&n,&k);
    Init();
    for (int i=1;i<=n;++i)
        scanf("%lld",&a[i]);
    sort(a+1,a+n+1);
    for (int i=1;i<=n;++i)
        ans=(ans+d[i-1]*a[i]%mod)%mod,ans=(ans-d[n-i]*a[i]%mod)%mod;;
    ans=(ans%mod+mod)%mod;
    printf("%lld\n",ans);
}