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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
ll modinv(ll a){
  ll b = mod,u = 1,v = 0,t;
  while(b){
    t = a/b;
    a -= t*b;swap(a,b);
    u -= t*v;swap(u,v);
  }
  u %= mod;
  if(u < 0) u += mod;
  return u;
}
int main(){
  ll N,K,i;scanf("%lld%lld",&N,&K);
  ll A[N],P[N];P[0] = P[1] = 1;
  for(i=0;i<N;i++) scanf("%lld",&A[i]);
  for(i=2;i<N;i++) P[i] = P[i-1]*i%mod;
  sort(A,A+N);ll ans = 0;
  for(i=0;i<=N-K;i++){
    ans = (ans-P[N-i-1]*modinv(P[K-1])%mod*modinv(P[N-i-K])%mod*A[i]%mod)%mod;
    ans = (ans+P[N-i-1]*modinv(P[K-1])%mod*modinv(P[N-i-K])%mod*A[N-i-1]%mod)%mod;
  }
  if(ans < 0) ans += mod;
  printf("%lld\n",ans);
}