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>

using namespace std;

typedef long long ll;

const ll M = 1e9 + 7;

ll n, k, ans;
ll a[100005];
ll fac[100005] = {1, 1};
ll inv[100005] = {1, 1};
ll fiv[100005] = {1, 1};

int main()
{
    scanf("%lld%lld", &n, &k);
    for(ll i = 0; i < n; i++)
    {
        scanf("%lld", &a[i]);
    }
    sort(a, a + n);
    for(ll 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(ll i = 0; i < n; i++)
    {
        ll j = n - i - 1;
        if(i >= k - 1) ans += a[i] * fac[i] % M * fiv[k - 1] % M * fiv[i - k + 1] % M;
        ans %= M;
        if(j >= k - 1) ans -= a[i] * fac[j] % M * fiv[k - 1] % M * fiv[j - k + 1] % M;
        ans += M;
        ans %= M;
    }
    printf("%lld", ans);
    return 0;
}
