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;
#define ll long long
#define lvector vector<ll>
#define rep(i,n) for(ll (i)=0; i<(n); ++(i))
#define print(a) cout << (a) << endl

const ll mod=1e9+7;

ll com(ll N, ll K, ll L) {
  ll d = 1, e = (L*K)%mod, m_ = mod-2;
  while(m_ != 0) {
    if (m_&1) d = (d*e)%mod;
    e = e*e%mod;
    m_ = m_ >> 1;
  }
  return (N*d)%mod;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n,k,val=1,ans=0;cin>>n>>k;
  lvector v(n,0),a(n,0);rep(i,n) cin>>v[i];
  sort(v.begin(),v.end());
  a[0]=1;
  for(ll i=1;i<=n-1;++i) a[i]=(a[i-1]*i)%mod;
  for(ll i=0;i<=n-k;++i) ans-=(v[i]*com(a[n-1-i],a[k-1],a[n-i-k]))%mod;
  for(ll i=k-1;i<n;++i) ans+=(v[i]*com(a[i],a[k-1],a[i-k+1]))%mod;
  ans%=mod;
  print(ans);
  return 0;
}