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 int N = 3e5 + 10;
const int MOD = 1e9 + 7;

ll qpow(ll x, ll k) {
    ll r = 1;
    while(k) {
        if(k & 1) r = r * x % MOD;
        k >>= 1;
        x = x * x % MOD;
    }
    return r;
}
ll d[N], a[N];
ll C(int i, int j) {
    return d[i] * qpow(d[i-j], MOD-2) % MOD * qpow(d[j], MOD - 2) % MOD;
}
int main() {
    d[0] = 1;
    for(int i = 1; i < N; i++) {
        d[i] = d[i-1] * i % MOD;
    }
    int n, k;
    cin >> n >> k;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(a, a + n);
    ll ans = 0;
    k--;
    for(int i = 0; i < n; i++) {
        int x = n - i - 1;
        if(x >= k) {
            ans -= C(x, k) * a[i] % MOD;
        }
        int y = i;
        if(y >= k) {
            ans += C(y, k) * a[i] % MOD;
        }
        ans = (ans + MOD) % MOD;
    }
    cout << ans << endl;
    return 0;
}