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;
typedef long long ll;
const ll mod = 1000000007;

const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];

void cominit()
{
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (ll i=2; i<MAX; i++) {
    fac[i] = fac[i-1] * i % mod;
    inv[i] = mod - inv[mod%i] * (mod/i) % mod;
    finv[i] = finv[i-1] * inv[i] % mod;
  }
}

ll com(ll n, ll k)
{
  if (n<k) return 0;
  if (n<0 || k<0) return 0;
  return fac[n] * (finv[k] * finv[n-k] % mod) % mod;
}

int main()
{
  ll N; cin >> N;
  ll K; cin >> K;
  vector<ll> A(N);
  for (int i=0; i<N; i++){
    cin >> A[i];
  }
  sort(A.begin(),A.end());
  cominit();

  ll p=0, m=0;
  for (int i=0; i<N; i++) {
    m += (A[i]*com(N-i-1, K-1))%mod;
    m %= mod;
    p += (A[i]*com(i, K-1))%mod;
    p %= mod;
  }
  cout << (p+mod-m)%mod << endl;

  return 0;
}
