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 <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()
#define ll long long
const ll MOD = 1000000007;
ll fact[100010];

ll modpow(ll x, ll y) {
	ll ans = 1;
	while (y) {
		if (y & 1)ans *= x;
		x *= x;
		ans %= MOD;
		x %= MOD;
		y /= 2;
	}
	return ans;
}

ll moddiv(ll x, ll y) {
	x %= MOD;y %= MOD;
	return x * modpow(y, MOD - 2) % MOD;
}

int main() {
	fact[0] = 1;
	for (ll i = 1;i < 100010;i++) {
		fact[i] = fact[i - 1] * i;
		fact[i] %= MOD;
	}
	ll N, K, A;
	cin >> N >> K;
	vector<ll> B(N);
	for (ll i = 0;i < N;i++) {
		cin >> B[i];
	}

	sort(all(B));

	ll ans = 0, b;
	for (ll i = 0;i < N;i++) {
		b = N - i - 1;
		if (b >= K - 1) {
			ans -= moddiv(fact[b], fact[K - 1] * fact[b - K + 1]) * B[i];
			ans %= MOD;ans += MOD;ans %= MOD;
		}
		if (i >= K - 1) {
			ans += moddiv(fact[i], fact[K - 1] * fact[i - K + 1]) * B[i];
			ans %= MOD;
		}
	}
	cout << ans << endl;
}