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 <iostream>
#include<bits/stdc++.h>
using namespace std;
long long a[100010];
long long fact[200020], finv[200020], inv[200020];
void facti() {
	fact[0] = fact[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < 200020; i++) {
		fact[i] = fact[i - 1] * i % 1000000007;
		inv[i] = 1000000007 - inv[1000000007 % i] * (1000000007 / i) % 1000000007;
		finv[i] = finv[i - 1] * inv[i] % 1000000007;
	}
}
long long ncr(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fact[n] * (finv[k] * finv[n - k] % 1000000007) % 1000000007;
}
int main() {
	long long n, k;
	cin >> n >> k;
	for (long long i = 0; i < n; i++) {
		cin >> a[i];
	}
	sort(a, a + n);
	facti();
	long long ans = 0;
	for (long long i = 0; i < n; i++) {
		ans = (ans - ncr(n - i - 1, k - 1) * a[i]) % 1000000007;
		ans = (ans + ncr(i, k - 1) * a[i]) % 1000000007;
	}
	cout << ans << endl;
	return 0;
}
