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<iostream>
#include<cmath>
#include<numeric>
#include<functional>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<iomanip>
#include<queue>
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<n;i++)
#define ll long long
#define INF 2147483647
using namespace std;
using Graph=vector<vector<int>>;

int main(){
  ll n,k; cin>>n>>k; ll a[n];
  rep(i,n){
    cin>>a[i];
  }
  sort(a,a+n);
  ll mod=1000000007;
  ll inv[n+1],hogeCkminus1[n]; inv[1]=1;
  for(int i=2;i<n+1;i++){
    inv[i]=mod-inv[mod%i]*(mod/i)%mod;
  }
  for(int i=0;i<k-1;i++){
    hogeCkminus1[i]=0;
  }
  hogeCkminus1[k-1]=1;
  for(int i=k;i<n;i++){
    hogeCkminus1[i]=(hogeCkminus1[i-1]*i)%mod;
    hogeCkminus1[i]=(hogeCkminus1[i]*inv[i-k+1])%mod;
  }
  // rep(i,n){cout<<hogeCkminus1[i]<<"\n";}
  ll ans=0;
  for(int i=0;i<n;i++){
    ans+=(a[i]*(hogeCkminus1[i]-hogeCkminus1[n-1-i]))%mod;
    ans%=mod;
  }
  cout<<ans;
}
