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;
const int mod = 1e9+7;
typedef long long LL;
int a[100005];
LL f[100005], r[100005];
LL inv(LL a){
    LL res = 1, b = mod-2;
    while(b){
        if(b%2 == 1) res = res*a%mod;
        a = a*a%mod;
        b/=2;
    }
    return res;
}
void prec(int n){
    f[0] = 1;
    for(int i=1;i<=n;i++) f[i] = f[i-1]*i%mod;
    for(int i=0;i<=n;i++) r[i] = inv(f[i]);
}
LL C(int a, int b){
    return f[a]*r[b]%mod*r[a-b]%mod;
}
int main(){
    int n, k;
    scanf("%d%d", &n, &k);
    prec(n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]);
    sort(a+1, a+1+n);
    for(int i=1;i<=n;i++) a[i] = (a[i] + mod)%mod;
    if(k == 1) printf("0\n");
    else{
        LL ans = 0;
        for(int i=k;i<=n;i++) ans = (ans + C(i-1, k-1)*a[i]%mod)%mod;
        for(int i=n-k+1;i>=1;i--) ans = (ans - C(n-i, k-1)*a[i]%mod + mod)%mod;
        printf("%lld\n", ans);
    }
}