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>
#define ll long long

using namespace std;

const int N = 1e5 + 100;
const int Mod = 1e9 + 7;

ll Fact[N];

ll FastPower (ll a, ll b) {
	if (!b) return 1;
	if (b & 1)	return a * FastPower (a, b - 1) % Mod;
	return FastPower (a * a % Mod, b >> 1);
}

ll C(int N, int R) {
	if (R > N) return 0;
	return (Fact[N] * FastPower((Fact[R] * Fact[N - R])%Mod, Mod - 2)) % Mod;
}

int main()
{
	Fact[0] = 1;
	for (int i = 1; i < N; i++)
		Fact[i] = i * Fact[i - 1] % Mod;

	int n, k;	scanf("%d%d", &n, &k);

	vector <int> Arr;
	for (int i = 0; i < n; i++) {
		int x;	scanf("%d", &x);
		Arr.push_back(x);
	}

	sort(Arr.begin(), Arr.end());

	ll Max = 0, Min = 0;
	for (int i = 0; i < n; i++) {
		Max = (Max + Arr[i] * C(i, k - 1) % Mod) % Mod;
		Min = (Min + Arr[i] * C(n - i - 1, k - 1) % Mod) % Mod;
	}

	cout << (Max - Min + Mod) % Mod << '\n';
	return 0;
}
