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 <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;
ll a[100009], pre[100009] = {0};
const ll MOD = 1e9+7;
ll qpow(ll a, ll b)
{
	ll res = 1;
	while(b)
	{
		if(b%2) res = res * a % MOD;
		a = a * a % MOD;
		b /= 2;
	}
	return res;
}
int main()
{
	int n, k;
	scanf("%d %d", &n, &k);
	for(int i = 1; i <= n; i++)
		scanf("%lld", a+i);
	
	sort(a+1, a+1+n);
	for(int i = 1; i <= n; i++)
		pre[i] = (pre[i-1] + a[i]) % MOD;
		
	if(k==1) printf("0\n");
	else
	{
		ll res = 0, tmp = 1;
		for(int i = k, j = k-2; i <= n; i++, j++)
		{
			ll sum = (pre[n] - pre[i-1]  - pre[n-i+1] + MOD * 2) % MOD;
			res = (res + tmp * sum % MOD) % MOD;
			tmp = (tmp * (j + 1) % MOD) * qpow(j-k + 3, MOD-2) % MOD;
		}
			printf("%lld\n", res);
	}
	
 
}