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;
typedef long long int ll;
ll arr[100005],p[100005],hell=pow(10,9)+7;
ll exp(ll x,ll y){
    ll res=1;
    while(y>0){
        if(y&1)
        res=(res*x)%hell;
        y>>=1;
        x=(x*x)%hell;
    }
    return res;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    p[0]=1;
    for(ll i=1;i<100005;i++)
    p[i]=(i*p[i-1])%hell;
    ll n,k,res=0;
    cin>>n>>k;
    for(ll i=0;i<n;i++)
    cin>>arr[i];
    sort(arr,arr+n);
    for(ll i=k-1;i<n;i++){
        ll sum=p[i];
        sum*=exp(p[i-k+1],hell-2);
        sum%=hell;
        res+=(sum*arr[i])%hell;
        res%=hell;
    }
    for(ll i=n-k;i>=0;i--){
        ll sum=p[n-i-1];
        sum*=exp(p[n-i-k],hell-2);
        sum%=hell;
        res-=(sum*arr[i])%hell;
        res%=hell;
    }
    res*=exp(p[k-1],hell-2);
    res%=hell;
    if(res<0)
    res+=hell;
    cout<<res;
}
