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;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int a[maxn], n, k;
ll norm(ll x)
{
	while(x < 0) x += mod;
	while(x >= mod) x -= mod;
	return x;
}
ll powMod(ll a, ll k)
{
	ll ret = 1;
	while(k)
	{
		if(k & 1) ret = ret * a % mod;
		a = a * a % mod; k >>= 1;
	}
	return ret;
}
ll getInv(ll x) { return powMod(x, mod - 2);}
ll solve()
{
	ll cur = 1, ret = 0;
	for(int i = k; i <= n; ++i)
	{
		ret += a[i] * cur % mod;
		ret = norm(ret);
		cur = cur *  i % mod * getInv(i - k + 1) % mod;
	}
	return ret;
}
int main()
{
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	sort(a + 1, a + n + 1);
	ll ret1 = solve();
	reverse(a + 1, a + n + 1);
	ll ret2 = solve();
	cout << norm(ret1 - ret2) << endl;
	return 0;
}