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
#define INF 1000000007ll
using namespace std;

int n,k;
int a[100010],pw[100010],rev[100010];

inline int my_pow(int x,int y)
{
  if(y==0) return 1;
  if(y==1) return x;
  int res=my_pow(x,y/2);
  if(y%2==0) return (res*res)%INF;
  else return (((res*res)%INF)*x)%INF;
}

signed main()
{
  ios::sync_with_stdio(false);
  cin>>n>>k;
  for(int i=1;i<=n;i++) cin>>a[i];
  sort(a+1,a+n+1);
  pw[0]=1;rev[0]=1;
  for(int i=1;i<=n;i++){
    pw[i]=(pw[i-1]*i)%INF;
    rev[i]=my_pow(pw[i],INF-2ll);
  }
  int ans=0ll;
  for(int i=1;i<=n-k+1;i++){
    int val=pw[n-i]*rev[n-i-k+1];
    val%=INF;
    val*=rev[k-1];val%=INF;
    val*=a[i];val%=INF;
    ans-=val;
  }
  for(int i=k;i<=n;i++){
    int val=pw[i-1]*rev[i-k];
    val%=INF;
    val*=rev[k-1];val%=INF;
    val*=a[i];val%=INF;
    ans+=val;
  }
  ans%=INF;
  if(ans<0) ans+=INF;
  cout<<ans<<endl;
  return 0;
}