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 long long long
using namespace std;
const int N=1e6,mod=1e9+7;
int aa[N+2];

int big(int b,int p,int m=mod)
{
  int ret=1;
  while(p){if(p&1)ret=(1LL*ret*b)%m;b=(1LL*b*b)%m;p/=2;}
  return ret;
}
int fac[N+2];
void init(int n)
{
  fac[0]=1;
  for(int i=1;i<=n;i++)fac[i]=(1LL*fac[i-1]*i)%mod;
}
int ncr(int n,int r)
{
  int a=fac[n];a=(1LL*a*big(fac[r],mod-2))%mod;
  return (1LL*a*big(fac[n-r],mod-2))%mod;
}

int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  int n,k;cin>>n>>k;
  for(int i=1;i<=n;i++)cin>>aa[i];
  sort(aa+1,aa+n+1,greater<int>());

  if(k==1)cout<<0<<endl,exit(0);

  // for(int i=1;i<=n;i++)cout<<aa[i]<<" ";
  //   cout<<endl;

  init(n);long ans=0;
  for(int i=1;i<=n;i++)
  {
    int rgt=n-i;
    if(rgt>=k-1)ans=(1LL*ans+1LL*ncr(rgt,k-1)*aa[i])%mod;

    int lft=i-1;
    if(lft>=k-1)ans=(1LL*ans-1LL*ncr(lft,k-1)*aa[i])%mod;

  }
  ans=(ans+mod)%mod;cout<<ans<<endl;
  
  return 0;
}