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.Arrays;
import java.util.Scanner;

public class Main {
	static long fact[] = new long[1000002];
	static long factInv[] = new long[1000002];

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		final int MOD = 1_000_000_000 + 7;

		fact[0] = 1;
		factInv[0] = 1;
		for(int i = 0; i < fact.length - 1; i++){
			fact[i+1] = fact[i]*(i+1) % MOD; // n!(mod M)
	        factInv[i+1] = (int)(factInv[i]*repeatablePow(i+1, MOD-2, MOD) % MOD);
		}

		int n = sc.nextInt();
		int k = sc.nextInt();

		int a[] = new int[n];
		for(int i = 0; i < n; i++){
			a[i] = sc.nextInt();
		}

		Arrays.sort(a);

		if(k == 1){
			System.out.println(0);
			return;
		}

		long sumArray[] = new long[n];
		sumArray[0] = a[0];
		for(int i = 1; i < n; i++){
			sumArray[i] = sumArray[i - 1] + a[i];
		}

		long result = 0;
		for(int i = k - 2; i <= n - 2; i++){
			result += (sumArray[n - 1] - sumArray[i] - sumArray[n - i - 2])%MOD*combination(i, k - 2, MOD);
			result %= MOD;
//			System.out.println("i = "+i+" result = "+result);
		}

		System.out.println(result);

	}

	static long repeatablePow(int x, int n, int mod){
		if(n == 0){
			return 1;
		}
		long result = repeatablePow((int)((long)x*x%mod), n/2, mod);
		if(n % 2 == 1){
			result = result*x%mod;
		}
		return result;
	}

	static long combination(int a, int b, int mod){ //aCbをmod計算
	    if(a == 0 && b == 0)return 1;
	    if(a < b || a < 0)return 0;

	    long tmp = factInv[a - b]* factInv[b] % mod;
	    return tmp * fact[a] % mod;
	}
}
