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 <algorithm>
#include <vector>

using namespace std;

const int MOD = 1000000007;
const int MAX = 110000;

long long fac[MAX], finv[MAX], inv[MAX];

void comInit() {
	fac[0] = fac[1] = inv[1] = finv[0] = finv[1] = 1;
	for (int i = 2; i < MAX; ++i) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

long long nCr(int n, int r) {
	if (n < r || n < 0 || r < 0) return 0;
	return fac[n] * (finv[r] * finv[n - r] % MOD) % MOD;
}

int main() {
	comInit();
	int n, k; cin >> n >> k;
	vector<int> a(n);
	for (auto& e : a) cin >> e;
	sort(a.begin(), a.end());
	long long smax = 0, smin = 0;
	for (int i = 0; i < n; ++i) {
		(smax += nCr(i, k - 1) * a[i]) %= MOD;
		(smin += nCr(n - i - 1, k - 1) * a[i]) %= MOD;
	}
	cout << (MOD + smax - smin) % MOD << endl;

	return 0;
}