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 <cstdio>
#include <algorithm>
using namespace std;
#define ll long long
ll mod = 1000000007;
ll big(ll b, ll p){
	if (p == 1) return b;
	ll h = big(b, p / 2);
	if (p % 2 == 0){
		return (h * h) % mod;
	}else{
		return (((h * h) % mod) * b) % mod;
	}
}
int main(){
	int n, k;
	scanf("%d %d", &n, &k);
	ll a[n];
	for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
	sort(a, a + n);
	ll x = 1, ans1 = 0, ans2 = 0, c = 1;
	for (int i = k - 1; i < n; i++){
		ans1 += a[i] * x;
		ans1 %= mod;
		x *= (i + 1);
		x %= mod;
		x *= big(c, mod - 2);
		x %= mod;
		c++;
	}
	reverse(a, a + n);
	x = 1;
	c = 1;
	for (int i = k - 1; i < n; i++){
		ans2 += a[i] * x;
		ans2 %= mod;
		x *= (i + 1);
		x %= mod;
		x *= big(c, mod - 2);
		x %= mod;
		c++;
	}
	printf("%lld\n", (ans1 - ans2 + mod) % mod);
}