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>
#define ll long long
using namespace std;
const int N = 1e5+1;
const ll mod = 1e9+7;
ll fact[N];
ll inv(ll a, ll p=mod-2) {
	if(p==0) return 1;
	ll b = inv(a, p/2);
	if(p%2==0) {
		return b*b%mod;
	}
	else {
		return b*b%mod*a%mod;
	}
}
ll C(int n, int k) {
	if(k>n) {
		return 0;
	}
	return fact[n]*inv(fact[k]*fact[n-k]%mod)%mod;
}
int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n, k;
	cin>>n>>k;
	fact[0]=1;
	for(ll i=1; i<=n; ++i) {
		fact[i]=fact[i-1]*i%mod;
	}
	vector<ll> a(n);
	for(int i=0; i<n; ++i) {
		cin>>a[i];
	}
	sort(a.begin(), a.end());
	reverse(a.begin(), a.end());
	ll ans=0;
	for(int i=0; i<n; ++i) {
		ll val = a[i] * C(n-i-1, k-1) % mod;
		ans += val;
		ans%=mod;
	}
	for(int i=1; i<=n; ++i) {
		ll val = a[n-i] * C(n-i, k-1) % mod;
		ans -= val;
		ans+=mod;
		ans%=mod;
	}
	cout<<ans<<'\n';
}
