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 <cstdio>
#include <algorithm>
typedef long long ll;
inline ll rd(){
    ll x=0,p=1;
    char a=getchar();
    while((a<48||a>57)&&a!='-')a=getchar();
    if(a=='-')p=-p,a=getchar();
    while(a>47&&a<58)x=(x<<1)+(x<<3)+(a&15),a=getchar();
    return x*p;
}
const int N=100002,S=100000;
const ll mod=1e9+7;
int n,k;
ll a[N],ans;
ll fac[N],inv[N];
inline ll fpow(ll b,ll p=mod-2){
	ll ans=1,tmp=b;
	while(p){
		if(p&1)ans=ans*tmp%mod;
		tmp=tmp*tmp%mod;
		p>>=1;
	}
	return ans;
}
inline ll C(int n,int m){
	return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int main(){
	n=rd(),k=rd();
	for(int i=1;i<=n;i++)a[i]=rd();
	std::sort(a+1,a+n+1);
	fac[0]=1;
	for(int i=1;i<=S;i++)fac[i]=fac[i-1]*i%mod;
	inv[S]=fpow(fac[S]);
	for(int i=S-1;i>=0;i--)inv[i]=inv[i+1]*(i+1)%mod;
	for(int i=1;i<=n;i++){
		if(i-1>=k-1)ans+=a[i]*C(i-1,k-1),ans%=mod;
		if(n-i>=k-1)ans-=a[i]*C(n-i,k-1),(ans+=mod)%=mod;
	}
	printf("%lld\n",ans);
    return 0;
}