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 MOD 1000000007
using ll=long long;
vector<ll> fact(200001);
void init_fact(int n){
	fact[0]=1;
	for (int i = 1; i < n+1; i++){
		fact[i]=fact[i-1]*i;
		fact[i]%=MOD;
	}
}
ll mod_pow(ll x, ll p){
	ll res=1;
	ll n=x;
	while(p>0){
		if(p&1==1)res=(res*n)%MOD;
		n=(n*n)%MOD;
		p=(p>>1);
	}
	return res;
}
ll nCr(int n, int r){
	ll res=1;
	res*=fact[n];
	res*=mod_pow(fact[r],MOD-2);
	res%=MOD;
	res*=mod_pow(fact[n-r],MOD-2);
	res%=MOD;
	return res;
}

int main() {
    int N,K;
    cin>>N>>K;
    init_fact(N+1);
    vector<ll> A(N);
    for (int i = 0; i < N; i++){
        cin>>A[i];
    }
    ll ans=0;
    sort(A.begin(),A.end());
    for (int i = N-1; i >= K-1; i--){
        ans+=(A[i]*nCr(i,K-1))%MOD;
        ans%=MOD;
    }
    for (int i = 0; i <= N-K; i++){
        ans-=(A[i]*nCr(N-i-1,K-1))%MOD;
        ans%=MOD;
    }
    ans=(ans+MOD)%MOD;
    cout<<ans<<endl;
}