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 Dukkha
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 100000, MD = 1000000007;

int aa[N], ff[N], gg[N];

int power(long long a, int k) {
	long long p = 1;
	while (k) {
		if (k & 1)
			p = p * a % MD;
		a = a * a % MD;
		k >>= 1;
	}
	return p;
}

long long ch(int n, int k) {
	return n < k ? 0 : (long long) ff[n] * gg[k] % MD * gg[n - k] % MD;
}

int main() {
	int n, k; cin >> n >> k;
	for (int f = 1, i = 0; i < n; i++) {
		gg[i] = power(ff[i] = f, MD - 2);
		f = (long long) f * (i + 1) % MD;
	}
	for (int i = 0; i < n; i++)
		cin >> aa[i];
	sort(aa, aa + n);
	long 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;
	cout << ans << '\n';
	return 0;
}
