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;

using ll = long long;
const ll M = 1e9 + 7;

ll power(ll a, ll b) {
  ll x = 1, y = a;
  while (b > 0) {
    if (b % 2) x = (x * y) % M;
    y = y * y % M;
    b /= 2;
  }
  return x % M;
}

ll inv(ll a) {
  return power(a, M - 2);
}

int main() {
  ll n, k; cin >> n >> k;
  vector<int> a(n);
  for (auto&x : a) cin >> x;
  
  sort(a.begin(),a.end());
  
  vector<ll> fac(n + 1);
  fac[0] = 1;
  for (ll i = 1; i <= n; i++) {
    fac[i] = (fac[i - 1] * i) % M;
  }
  
  auto comb = [&](ll a, ll b) -> ll {
    ll res = fac[a];
    res = (res * inv(fac[b])) % M;
    res = (res * inv(fac[a - b])) % M;
    return res;
  };
  
  ll ans = 0;
  for (int i = n - 1; i >= k - 1; i--) {
    ll tmp = (comb(i, k - 1) * a[i]) % M;
    ans = (ans + tmp) % M;
    
    tmp = (comb(i, k - 1) * a[n - 1 - i]) % M;
    ans = (ans - tmp + M) % M;
  }
  cout << (ans+M)%M << endl;
}