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 vector<int> VI;
typedef pair <int,int> ii;
typedef long long LL;
#define pb push_back
const int INF = 2147483647;
const int MOD = 1000000007;
const int N = 500005;

int n, k, tab[N], i, res, act;

int inverse(int a, int n = MOD - 2) {
    int r = 1;
    while (n) {
        if (n % 2) r = (r * 1LL * a) % MOD;
        a = (a * 1LL * a) % MOD;
        n >>= 1;
    }
    return r;
}

int main() {
scanf("%d %d", &n, &k);
for (i=1;i<=n;i++) scanf("%d", &tab[i]);
sort(tab + 1, tab + n + 1);
act = 1;
for (i=k;i<=n;i++) {
	res = (tab[i] * 1LL * act + res) % MOD;
	res = (-tab[n - i + 1] * 1LL * act + res) % MOD;
	if (i < n) {
		act = (act * 1LL * i) % MOD;
		act = (act * 1LL * inverse(i - k + 1)) % MOD;
	}
}
if (res < 0) res += MOD;
printf("%d\n", res);
}