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;
#define pb push_back
#define mp make_pair
long long n,k,arr[100001],fact[100001];
long long total,p;
long long power(long long x,long long y){
    long long res=1;
    while (y>0){
        if (y%2==1) res=(res*x)%p;
        y/=2;
        x=(x*x)%p;
    }
    return res;
}
long long ncr(long long x,long long y){
    if (x<y) return 0;
    long long ans=fact[x];
    ans*=power(fact[y],p-2);
    ans%=p;
    ans*=power(fact[x-y],p-2);
    ans%=p;
    return ans;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>k;
    for (int i=1;i<=n;i++) cin>>arr[i];
    p=1e9+7;
    sort(arr+1,arr+n+1);
    fact[0]=1;
    for (int i=1;i<=n;i++){
        fact[i]=fact[i-1]*i;
        fact[i]%=p;
    }
    for (int i=1;i<=n;i++){
        total+=arr[i]*ncr(i-1,k-1);
        total%=p;
        total-=arr[i]*ncr(n-i,k-1);
        total%=p;
    }
    cout<<total;
}