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 rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
typedef pair<int,int> P;
typedef vector<int> vec;
typedef vector<vec> mat;
ll mod=1e9+7;
ll inv(ll a){
    ll n=mod-2,ans=1;
    while(n>0){
        if(n&1) ans=ans*a%mod;
        n=n/2;
        a=a*a%mod;
    }
    return ans;
}
ll A[200001];
void init(){
    A[0]=1;
    rep(i,200000) A[i+1]=A[i]*(i+1)%mod;
}
ll nCk(ll n,ll k){
    return A[n]*inv(A[k])%mod*inv(A[n-k])%mod;
}
int main(){
    init();
    ll n,k,B[200001];
    cin>>n>>k;
    rep(i,n) cin>>B[i];
    sort(B,B+n);
    ll ma=0,mi=0;
    rep(i,n){
        if(i>=k-1) ma=(ma+B[i]*nCk(i,k-1))%mod;
        if(i<=n-k) mi=(mi+B[i]*nCk(n-i-1,k-1))%mod;
    }
    cout<<(ma-mi+mod)%mod<<"\n";
}