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 <cstdio>
#include <algorithm>
using namespace std;
const long long mod = 1000000007;
long long mp(long long b, long long e) {
	if (e == 0) return 1;
	long long a = mp(b, e / 2);
	a = a * a % mod;
	if (e % 2) a = a * b % mod;
	return a;
}
int main() {
	int N, K;
	scanf("%d %d", &N, &K);
	long long A[N];
	for (int i = 0; i < N; ++i) scanf("%lld", &A[i]);
	sort(A, A + N);
	long long ans = 0, fac[N + 1], inv[N + 1];
	fac[0] = inv[0] = 1;
	for (int i = 1; i <= N; ++i) {
		fac[i] = i * fac[i - 1] % mod;
		inv[i] = mp(fac[i], mod - 2);
	}
	for (int i = 0; i <= N - K; ++i) ans = (ans - fac[N - 1 - i] * inv[K - 1] % mod * inv[N - i - K] % mod * A[i] % mod) % mod;
	for (int i = K - 1; i < N; ++i) ans = (ans + fac[i] * inv[K - 1] % mod * inv[i - K + 1] % mod * A[i] % mod) % mod;
	printf("%lld", (ans + mod) % mod);
	return 0;
}