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 ll long long
const ll mod=1e9+7;

ll bigmod(ll b,ll p)
{
    if(!p)return 1;
    ll x=bigmod(b,p/2);
    x=(x*x)%mod;
    if(p&1)x=(x*b)%mod;
    return x;
}
int main()
{
    ll n,k;cin>>n>>k;
    vector<ll>vc;
    for(int i=1;i<=n;i++)
    {
        ll x;cin>>x;
        vc.push_back(x);
    }
    sort(vc.begin(),vc.end());

    ll fact[n+3],cal[n+3];
    fact[0]=1;
    for(ll i=1;i<=n+1;i++)fact[i]=(fact[i-1]*i)%mod;
    for(ll i=k-1;i<=n;i++)
    {
        ll ret=(fact[k-1]*fact[i-k+1])%mod;
        cal[i]=(fact[i]*bigmod(ret,mod-2))%mod;
    }
    ll sum=0;

    for(ll i=n-1;i>=k-1;i--)
    {
       ll ret=(vc[i]*cal[i])%mod;
       sum=(sum+ret)%mod;
    }
   // cout<<sum<<"\n";

    for(ll i=0;i<=n-k;i++)
    {
         ll ret=(vc[i]*cal[n-i-1])%mod;
       sum=((sum-ret)%mod+mod)%mod;
    }
    cout<<sum<<"\n";
}