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 ll MOD = 1e9+7;
const int N = 1e5;

ll dmod(ll x){
	if(x < 0){
		return MOD - ((-x) % MOD);
	}
	return x%MOD;
}

ll fe(ll x, ll p){
	if(p==0) return 1ll;
	if(p==1) return x;
	ll ret = fe(x,p/2);
	ret = dmod(ret*ret);
	if(p&1) ret = dmod(ret*x);
	return ret;
}

int n,k;
ll li[N+5];
ll fakto[N+5];

ll nCr(ll n, ll r){
	ll inv = fe(dmod(fakto[r] * fakto[n-r]),MOD-2);
	return dmod(fakto[n] * inv);
}

int main(){
	fakto[0] = 1ll;
	for(int i=1;i<=N;i++){
//		cout << fakto[i-1] << endl;
		fakto[i] = dmod(fakto[i-1] * (ll)(i));
	}
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++) scanf("%lld",&li[i]);
	sort(li+1,li+n+1);
	ll ans = 0ll;
	for(int i=1;i<=n-k+1;i++){ //minimum
		ans = dmod(ans - dmod(nCr(n-i,k-1) * li[i]) + MOD);
	}
	for(int i=n;i>=k;i--){
		ans = dmod(ans + dmod(nCr(i-1,k-1) * li[i]));
	}
	printf("%lld\n",ans);
	return 0;
}
