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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int mod = 1000000007;
int binpow(int a, int b) {
	int ans = 1;
	while (b > 0) {
		if (b & 1) ans = 1LL * ans * a % mod;
		a = 1LL * a * a % mod;
		b >>= 1;
	}
	return ans;
}
int solve(int N, int K, vector<int> A) {
	int way = 1, ans = 0;
	for (int i = K - 1; i < N; ++i) {
		int sub = 1LL * way * ((A[i] >= 0 ? A[i] : mod - (-A[i]) % mod)) % mod;
		ans = (ans + sub) % mod;
		way = 1LL * way * (i + 1) % mod;
		way = 1LL * way * binpow(i - K + 2, mod - 2) % mod;
	}
	return ans;
}
int main() {
	int N, K;
	cin >> N >> K;
	vector<int> A(N);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
	}
	sort(A.begin(), A.end());
	int res1 = solve(N, K, A);
	reverse(A.begin(), A.end());
	int res2 = solve(N, K, A);
	cout << (res1 - res2 + mod) % mod << endl;
	return 0;
}