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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int n = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc.
        int m = in.nextInt();
        in.nextLine();

        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
        }
        System.out.println(getMax(arr, m));
    }

    private static int getMax(int[] arr, int m) {
        Arrays.sort(arr);
        long sum = 0;
        int mod = 1000000007;
        int l = arr.length;
        long base = 1;

        for (int i = 0; i <= l - m; i++) {
            //System.out.println(arr[m + i - 1] + " " + arr[l - m - i] + " " + base);
            sum = (sum + base * arr[m + i - 1] - base * arr[l - m - i]) % mod;
            base = (base * (m + i) % mod) * ((getInverseMod(i + 1, mod)[0] + mod) % mod) % mod;
        }

        return (int) ((sum + mod) % mod);
    }

    private static long[] getInverseMod(int i, int mod) {
        if (mod == 0) {
            return new long[]{1, 0};
        }
        long[] temp = getInverseMod(mod, i  % mod);
        return new long[]{temp[1], temp[0] - i / mod * temp[1]};
    }
}
