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;
#define mod (1000000007)
const int MAX=100010;
ll n,k,a[MAX],ans=0;
ll fact[MAX];
ll fact_inv[MAX];

ll power(ll x,ll n)
{
    ll res=1;
    while(n){
        if(n&1)res=res*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return res;
}
void init(){
    fact[0]=1;
    for(int i=0;i<n;i++)fact[i+1]=fact[i]*(i+1)%mod;
    fact_inv[n]=power(fact[n],mod-2);
    for(int i=n-1;i>=0;i--)fact_inv[i]=fact_inv[i+1]*(i+1)%mod;
}
ll comb(ll n,ll r){
    if(n<0||r<0||n<r)return 0;
    return (fact[n]*fact_inv[r])%mod*fact_inv[n-r]%mod;
}
void solve() {
    cin>>n>>k;
    for(int i=0;i<n;i++)cin>>a[i];
    sort(a,a+n);
    init();
    for(int i=0;i<n;i++)ans-=a[i]*comb(n-i-1,k-1),ans=(ans+a[i]*comb(i,k-1))%mod;
    cout<<ans<<endl;
}
int main()
{
    cin.tie(0);
	ios::sync_with_stdio(false);
    solve();
    return 0;
}
