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;

long long MOD = 1000000007;

vector<long long> F, RF;

void init(long long N) {
	F.resize(N + 1), RF.resize(N + 1);
	vector<long long> inv(N + 1);
	inv[1] = 1;
	F[0] = F[1] = RF[0] = RF[1] = 1;
	for (int i = 2; i <= N; i++) {
		F[i] = (F[i - 1] * i) % MOD;
		inv[i] = MOD - (inv[MOD % i] * (MOD / i)) % MOD;
		RF[i] = (RF[i - 1] * inv[i]) % MOD;
	}
}

long long Comb(long long A, long long B) {
	if (B < 0 || A < B) return 0;
	return (F[A] * ((RF[A - B] * RF[B]) % MOD)) % MOD;
}

int main() {
	long long N, K, MAXS = 0, MINS = 0;
	cin >> N >> K;
	init(N);
	vector<long long> A(N);
	for (int i = 0; i < N; i++) cin >> A[i];
	sort(A.begin(), A.end());
	for (int i = 0; i < N; i++) {
		MAXS = (MAXS + Comb(i, K - 1) * A[i]) % MOD;
	}
	reverse(A.begin(), A.end());
	for (int i = 0; i < N; i++) {
		MINS = (MINS + Comb(i, K - 1) * A[i]) % MOD;
	}
	cout << (MOD + MAXS - MINS) % MOD << endl;
}