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 <cstdio>
#include <algorithm>
#define N 100005
using namespace std;
typedef long long ll;

ll n, k, ans, M=1e9+7, a[N];
ll fac[N]={1,1}, inv[N]={1,1}, fiv[N]={1,1};
ll c(ll p, ll q) {return fac[p] * fiv[q] % M * fiv[p-q] % M;}

int main() {
	ll i, t;
	cin >> n >> k;
	for(i=0; i<n; i++) scanf("%lld", &a[i]);
	sort(a, a+n);
	for(i=2; i<=n; i++) {
		fac[i] = fac[i-1] * i % M;
		inv[i] = inv[M%i] * (M-M/i) % M;
		fiv[i] = fiv[i-1] * inv[i] % M;
	}
	for(i=k-1; i<n; i++) {
		(ans += a[i] * c(i, k-1)) %= M;
		(ans -= a[n-i-1] * c(i, k-1)) %= M;
	}
	cout<<ans;
	return 0;
}