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;
#define maxn 100005
#define mod 1000000007
typedef long long LL;
LL c[maxn] = {};
LL modinv(LL a, LL p) {
    if (p == 1) return 0;
    LL pp = p, y = 0, x = 1;
    while (a > 1) {
        LL q = a / p, t = p;
        p = a % p, a = t, t = y, y = x - q * y, x = t;
    }
    if (x < 0) x += pp;
    return x;
}
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	LL a[maxn];
	LL sum[maxn] = {};
	for(int i=1; i<=n; i++)
		cin >> a[i];
	sort(a+1, a+n+1);
	for(int i=1; i<=n; i++)
	{
		sum[i] = (sum[i-1] + a[i]) % mod;
	}
	if(k<2)
	{
		cout << 0 << endl;
		return 0;
	}
	k -= 2;
	c[k] = 1;
	for(int i=k+1; i<=n; i++)
	{
		c[i] = c[i-1] * i % mod * modinv(i-k, mod) % mod;
	}
	LL ans = 0;
	for(int i=k; i<=n-2; i++)
	{
		ans = (ans + (sum[n]-sum[n-i-1]-sum[i+1]) % mod * c[i] % mod) % mod;
	}
	cout << ans << "\n";
	return 0;
}