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;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=1e9+7;

ll power(ll x,ll n){
    if(n==0) return 1;
    ll res=power(x*x%mod,n/2);
    if(n&1) res=res*x%mod;
    return res;
}

int main(){
  int n,k;
  cin>>n>>k;
  ll c[n];
  rep(i,0,n) cin>>c[i];
  sort(c,c+n);
  
  ll a[n+1];
  a[0]=1;
  rep(i,1,n+1) a[i]=a[i-1]*i%mod;
  ll b[n+1];
  b[0]=1;
  rep(i,1,n+1) b[i]=power(a[i],1e9+5);
  
  ll cur=0;
  ll ans=0;
  rep(i,0,n-k+1){
    cur+=c[n-1-i]-c[i];
    ans+=cur%mod*a[n-2-i]%mod*b[k-2]%mod*b[n-k-i]%mod;
    ans%=mod;
  }
  cout<<ans<<endl;
}
