#include "bits/stdc++.h"
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
const int M=100010;
int INF=1000000007;
using ll=long long;

ll C[51][51];
void comb_table(int N){
	rep(i,0,N+1){
		rep(j,0,i+1){
			if(j==0||j==i)C[i][j]=1LL;
			else C[i][j]=(C[i-1][j-1]+C[i-1][j]);
		}
	}
}

int main(){
	int n,a,b;
	double ave=0.0;
	vector<ll> v(55);
	cin>>n>>a>>b;
	rep(i,0,n)cin>>v[i];
	comb_table(n);
	sort(ALL(v),greater<ll>());
	rep(i,0,a)ave+=v[i];
	ave/=a;
	//v[a-1]と同じ値を持つ数の個数をl,
	//大きい方から,v[a-1]を使った個数をcur_lとする.
	int l=0,cur_l=0;
	rep(i,0,n){
		if(v[i]==v[a-1]){
			l++;
			if(i<a)
				cur_l++;
		}
	}
	ll cnt=0;
	//大きい方から並べてa個までの数が全て同じ時.
	if(cur_l==a)
		rep(i,a,b+1)cnt+=C[l][i];
	//そうでないとき(最後に使った文字の総数)C(最後に使った文字の使った文字数);
	else cnt+=C[l][cur_l];
	cout.precision(20);
	cout<<fixed<<ave<<endl;
	cout<<cnt<<endl;;
}
### Prompt

Create a solution in Cpp for the following problem:

You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
const int M=100010;
int INF=1000000007;
using ll=long long;

ll C[51][51];
void comb_table(int N){
	rep(i,0,N+1){
		rep(j,0,i+1){
			if(j==0||j==i)C[i][j]=1LL;
			else C[i][j]=(C[i-1][j-1]+C[i-1][j]);
		}
	}
}

int main(){
	int n,a,b;
	double ave=0.0;
	vector<ll> v(55);
	cin>>n>>a>>b;
	rep(i,0,n)cin>>v[i];
	comb_table(n);
	sort(ALL(v),greater<ll>());
	rep(i,0,a)ave+=v[i];
	ave/=a;
	//v[a-1]と同じ値を持つ数の個数をl,
	//大きい方から,v[a-1]を使った個数をcur_lとする.
	int l=0,cur_l=0;
	rep(i,0,n){
		if(v[i]==v[a-1]){
			l++;
			if(i<a)
				cur_l++;
		}
	}
	ll cnt=0;
	//大きい方から並べてa個までの数が全て同じ時.
	if(cur_l==a)
		rep(i,a,b+1)cnt+=C[l][i];
	//そうでないとき(最後に使った文字の総数)C(最後に使った文字の使った文字数);
	else cnt+=C[l][cur_l];
	cout.precision(20);
	cout<<fixed<<ave<<endl;
	cout<<cnt<<endl;;
}
```