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 maxn=1e5+10,mod=1e9+7;
int n,k,a[maxn];
ll Min,Max;
ll fac[maxn],inv[maxn];
ll fpm(ll a,ll b){
    ll ret=1ll;
    for(;b;b>>=1,a=a*a%mod)
        if(b&1)ret=ret*a%mod;
    return ret;
}
void init(){
    fac[0]=inv[0]=1;
    for(int i=1;i<=n;++i){
        fac[i]=fac[i-1]*i%mod;
        inv[i]=fpm(fac[i],mod-2);
    }
}
ll C(ll n,ll k){
    if(n<0||k<0||n-k<0)return 0;
    return fac[n]*inv[k]%mod*inv[n-k]%mod;
}

void solve(){
    ios::sync_with_stdio(false);
    cin>>n>>k;
    init();
    for(int i=1;i<=n;++i){
        cin>>a[i];
    }
    sort(a+1,a+n+1);
    for(int i=1;i<=n;++i){
        Min+=1ll*(a[i]+mod)*C(n-i,k-1)%mod;
        Min%=mod;
    }
    sort(a+1,a+n+1,greater<int>());
    for(int i=1;i<=n;++i){
        Max+=1ll*(a[i]+mod)*C(n-i,k-1)%mod;
        Max%=mod;
    }
    cout<<(Max-Min+mod)%mod<<endl;
}
int main(){
    solve();
    return 0;
}