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 int long long
using namespace std;
const int mod=1000000007;
int k[100100];
int p(int A,int B){
  if(!B)return 1;
  if(B%2)return p(A,B-1)*A%mod;
  int C=p(A,B/2);
  return C*C%mod;
}
int c(int A,int B){
  if(A<B)return 0;
  return k[A]*p(k[A-B]*k[B]%mod,mod-2)%mod;
}
signed main(){
  k[0]=1;
  for(int i=1;i<=100000;i++)k[i]=k[i-1]*i%mod;
  int N,K;
  cin>>N>>K;
  vector<int> A(N);
  for(int &i:A)cin>>i;
  sort(A.begin(),A.end());
  int ans=0;
  for(int i=0;i<N;i++)ans=(ans+A[i]*((c(i,K-1)-c(N-i-1,K-1)+mod)%mod)%mod+mod)%mod;
  cout<<ans<<endl;
}