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 mod = 1000000007;
const int MAXN = 100005;
LL num[MAXN];
LL fact[MAXN];
LL my_pow(LL a, LL b)
{
    LL sum=1;
    while(b>0)
    {
        if(b&1)
            sum=sum*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return sum%mod;
}
void init()
{
    fact[0]=1;
    for(int i=1;i<MAXN;++i)
       fact[i]=fact[i-1]*i%mod;
}
LL comb(LL a, LL b)
{
    if(b>a)
        return 0;
    return fact[a] *  my_pow(fact[b] * fact[a-b] % mod, mod-2) % mod;
}
int main()
{
   int n, k;
   init();
   scanf("%d %d", &n, &k);
   for(int i=0;i<n;++i)
       scanf("%lld", &num[i]);
   LL ans=0;
   sort(num, num+n);
   for(int i=0;i<n-k+1;++i)
   {
       ans = (ans + (num[n-i-1]-num[i])%mod*comb(n-i-1,k-1)%mod)%mod;
       //printf("%d %d %lld\n", n-i-1, k-1, (num[n-i-1]-num[i])*comb(n-i-1,k-1)%mod);
   }
   printf("%lld\n" ,(ans % mod + mod)%mod);
   return 0;
}