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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
using ll = long long;
#define repl(i, l, r) for (ll i = l; i < r; i++)
#define rep(i, n) repl(i, 0, n)
const ll MOD = 1000000007;
const ll MAX = 1100000;

ll fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	repl(i, 2, MAX) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}
ll COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 or k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int main() {
	COMinit();
	ll n, k;
	cin >> n >> k;
	ll a[n];
	rep(i, n) cin >> a[i];
	sort(a, a + n);
	ll M = 0, m = 0;
	rep(i, n) {
		M += a[i] * COM(i, k - 1);
		m += a[i] * COM(n - 1 - i, k - 1);
		M %= MOD, m %= MOD;
	}
	cout << (M - m + MOD) % MOD << endl;
	return 0;
}