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 <algorithm>
#include <iostream>

using namespace std;

const int M = 1e9+7;

int mpow(long long a, int n)
{
  if (n % 2) return mpow(a, n-1) * a % M;
  if (n == 0) return 1;
  return mpow(a * a % M, n/2);
}

int main()
{
  int n, k; cin >> n >> k;
  int a[100005]; for (int i = 0; i < n; ++i) cin >> a[i];
  
  sort(a, a+n);
  
  long long kaijo[100005]; kaijo[0] = 1;
  for (int i = 1; i <= n; ++i) kaijo[i] = kaijo[i-1] * i % M;
  
  int inv[100005];
  for (int i = 0; i <= n; ++i) inv[i] = mpow(kaijo[i], M-2);
  
  long long com[100005];
  for (int i = k-1; i <= n; ++i) com[i] = kaijo[i] * inv[k-1] % M *inv[i-k+1] % M;
  int mn = 0, mx = 0;
  for (int i = 0; i < n-k+1; ++i) {
    mn = (mn + (a[i] + M) * com[n-i-1] % M) % M;
    mx = (mx + (a[n-i-1] + M) * com[n-i-1] % M) % M;
  }
  
  int ans = (mx - mn + M) % M;
  cout << ans << endl;
  return 0;
}