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 pb push_back
const int N = 100005;
# define MOD 1000000007
typedef long long i64;

int a[N];
i64 A[N];

i64 pow_mod(i64 a, i64 n, i64 mod){
    i64 ret=1, tmp=a%mod;
    while (n) {
        if (n&1) ret=ret*tmp%MOD;
        tmp=tmp*tmp%MOD;
        n>>=1;
    }
    return ret;
}
i64 inv(i64 a, i64 mod){return pow_mod(a,mod-2,mod);}
i64 C(int x, int y) {
    return A[x]*inv(A[y], MOD)%MOD*inv(A[x-y], MOD)%MOD;
}
int main ()
{
    int n, k;
    cin >> n >> k;
    A[0] = 1;
    for (int i = 1; i <= n; ++i) A[i] = A[i-1]*i%MOD;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    sort(a + 1, a + n + 1);
    i64 ans = 0;
    for (int i = 1; i <= n-k+1; ++i) {
        ans = (ans - a[i] * C(n-i, k-1) % MOD)%MOD;
    }
    for (int i = n; i >= k; --i) {
        ans = (ans + a[i] * C(i-1, k-1) % MOD)%MOD;
    }
    cout << (ans + MOD)%MOD << endl;
    return 0;
}
