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
// practice with kaiboy
import java.io.*;
import java.util.*;

public class Main extends PrintWriter {
	Main() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		Main o = new Main(); o.main(); o.flush();
	}

	static final int MD = 1000000007;
	int power(long a, int k) {
		long p = 1;
		while (k > 0) {
			if ((k & 1) == 1)
				p = p * a % MD;
			a = a * a % MD;
			k >>= 1;
		}
		return (int) p;
	}
	int[] ff, gg;
	long ch(int n, int k) {
		return n < k ? 0 : (long) ff[n] * gg[k] % MD * gg[n - k] % MD;
	}
	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		ff = new int[n];
		gg = new int[n];
		for (int f = 1, i = 0; i < n; i++) {
			gg[i] = power(ff[i] = f, MD - 2); 
			f = (int) ((long) f * (i + 1) % MD);
		}
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		aa = Arrays.stream(aa).boxed().sorted().mapToInt($->$).toArray();
		long ans = 0;
		for (int i = 0; i < n; i++)
			ans = (ans + (aa[i] - aa[n - 1 - i]) * ch(i, k - 1)) % MD;
		if (ans < 0)
			ans += MD;
		println(ans);
	}
}
