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;
typedef long long ll;
const int N=1e5+10;
  int n,mod=1e9+7;
ll fact[N],inv[N],arr[N];
 ll fp( ll base,ll exp){
 if(exp==0)
 return 1;
 ll ans=fp(base,exp/2);
 ans=(ans*ans)%mod;
 if(exp%2!=0){
    ans=(ans*(base%mod))%mod;
 }
    return ans;
 }
ll ncr(ll n, ll r)
{
    return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;
}
int main()
{
  //cout<< fp(5,2);

     fact[0]=1;
     inv[0]=1;
   for(int i=1;i<=1e5+5;i++){
    fact[i]=(fact[i-1]*i)%mod;
    inv[i] = fp(fact[i], mod - 2);

   }

   ll ans=0;
   int k;
   cin>>n>>k;
   for(int i=0;i<n;i++){
    cin>>arr[i];
   }
   sort(arr,arr+n);
   for(int i=0;i<n;i++){
        if(n-i-1>=k-1&&k-1!=0){
            ans-=(ncr(n-i-1,k-1)*arr[i])%mod;
            ans%=mod;
        }

    if(i>=k-1&&k-1!=0){
         ans+=(ncr(i,k-1)*arr[i])%mod;
         ans%=mod;
    }


   }
cout<<ans;

   return 0;
}
