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=1e9+7;
ll ans;
ll fac[100005];
ll sq(ll a){
	return (a*a)%mod;
}
ll fp(ll a,ll b){
	if(b==0)return 1;
	if(b==1)return a;
	if(b&1){
		return (sq(fp(a,b>>1))*a)%mod;
	}
	return sq(fp(a,b>>1))%mod;
}
ll cnk(ll n,ll k){
	if(n<k)return 0;
	if(n==k)return 1;
	if(k==0)return 1;
	ll res=fac[n];
	res*=fp(fac[k],mod-2);
	res%=mod;
	res*=fp(fac[n-k],mod-2);
	return (res)%mod;
}
int main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
	int n,k;cin>>n>>k;
	vector<ll>v;
	for(int i=0;i<n;i++){
		ll a;cin>>a;
		v.push_back(a);
	}
	if(k==1){
		cout<<0<<endl;return 0;
	}
	sort(v.begin(),v.end());
	fac[0]=1;
	for(int i=1;i<1e5;i++){
		fac[i]=fac[i-1]*i;
		fac[i]%=mod;
	}
	for(int i=0;i<n;i++){
		ans+=v[i]*cnk(i,k-1);
		ans%=mod;
		ans-=v[i]*cnk(n-i-1,k-1);
		ans%=mod;ans+=mod;ans%=mod;
	}
	cout<<ans;
	return 0;
}