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
import java.util.*;
import java.util.function.*;
 
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int K = sc.nextInt();
    long[] A = new long[N];
    for (int i = 0; i < N; i++) A[i] = sc.nextLong();
    sc.close();

    long mod = 1000000007L;
    long[] cmb = new long[N - K + 1];
    cmb[0] = 1;
    for (int i = 1; i <= N - K; i++)
      cmb[i] = modDiv(cmb[i - 1] * (K + i - 1), i, mod);

    Arrays.sort(A);
    long ans = 0;
    for (int i = 0; i < N - K + 1; i++) {
      ans -= (A[i] * cmb[N - K - i]) % mod;
      ans %= mod;
      ans += (A[N - 1 - i] * cmb[N - K - i]) % mod;
      ans %= mod;
    }

    if (ans < 0) ans += mod;
    System.out.println(ans);
  }

  private static long modDiv(long a, long b, long mod) {
    return (a % mod) * modInv(b, mod) % mod;
  }

  private static long modInv(long x, long mod) {
    return modPow(x, mod - 2, mod);
  }

  private static long modPow(long b, long e, long mod) {
    long ans = 1;
    long x = b % mod;
    for (long y = e; y != 0; y >>= 1) {
      if ((y & 1) == 1) ans = (ans * x) % mod;
      x = (x * x) % mod;
    }
    return ans;
  }
}