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;

const int base = 1000000007;

int inv(long long p)
{
  int q = base - 2, r = 1;
  while (q)
  {
    if (q & 1) r = r * p % base;
    
    p = p * p % base;
    q >>= 1;
  }
  
  return r;
}

int main()
{
  int n, k; cin >> n >> k;
  if (k == 1) { cout << 0 << endl; return 0; }
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  
  sort(a.begin(), a.end());
  for (int i = 1; i < n; ++i)
    a[i] = (a[i] + a[i-1]) % base;
  
  long long combi = 1;
  long long total = 0;
  for (int i = k; i <= n; ++i)
  {
    int total_i = ( (a[n-1] - a[i-2]) % base - a[n-i] ) % base * combi % base;
    total = (total + total_i) % base;
    
    combi = combi * (i-1) % base * inv(i-k+1) % base;
  }
  
  if (total < 0) total += base;
  cout << total << endl;
}