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 <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
using namespace std;

const int mod = 1e9 + 7;;
const int N = 1e5 + 10;
long long num[N];
long long result[N];
long long power(long long a, long long b) {
	long long res = 1;
	while (b > 0) {
		if (b & 1)
			res = res * a % mod;
		a = a * a % mod;
		b >>= 1;
	}

	return res % mod;
}

long long comb(long long a, long long b) {
	if (b > a)
		return 0;
	return result[a] * power(result[b] * result[a - b] % mod, mod - 2) % mod;
}

int main() {
	int n, k;
	result[0] = 1;
	for (int i = 1; i < N; ++i)
		result[i] = result[i - 1] * i % mod;
	
	cin >> n >> k;
	for (int i = 0; i < n; ++i)
		cin >> num[i];
	long long ans = 0;

	sort(num, num + n);

	for (int i = 0; i < n - k + 1; i++)
		ans = (ans + (num[n - i - 1] - num[i]) % mod * comb(n - i - 1, k - 1) % mod) % mod;
	cout << ((ans % mod + mod) % mod) << endl;
	return 0;
}