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>
#define rp(i,n) for(int i=0; i<n; i++)
using namespace std;
using ll=long long;
typedef pair<int,int> P;

ll RepeatSquare(ll x,ll y,ll mod){
    if(y==0) return 1;
    if(y%2==0){
        ll a= RepeatSquare(x,y/2,mod);
        return a*a%mod;
    }
    if(y%2==1) {
        return x*RepeatSquare(x,y-1,mod)%mod;
    }
}

int main(){
    ll n,k;
    cin >> n >> k;
    vector<ll> a(n);
    rp(i,n) cin >> a.at(i);
    sort(a.begin(),a.end());
    ll com;
    ll ans=0;
    for(ll i=k; i<=n;i++){
        if(i==k) com=1;
        else{
            com*=(i-1);
            com%=1000000007;
            com*=RepeatSquare(i-k,1000000005,1000000007);
            com%=1000000007;
        }
        ans+=(a.at(i-1)-a.at(n-i))*com;
        ans%=1000000007;

    }
    cout << ans << endl;
    return 0;
}