Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

5 2 2
1 2 3 4 5

Sample Output 1

4.500000
1

The mean of the values of selected items will be maximized when selecting the fourth and fifth items. Hence, the first line of the output should contain 4.5.

There is no other way to select items so that the mean of the values will be 4.5, and thus the second line of the output should contain 1.

Sample Input 2

4 2 3
10 20 10 10

Sample Output 2

15.000000
3

There can be multiple ways to select items so that the mean of the values will be maximized.

Sample Input 3

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996

Sample Output 3

1000000000000000.000000
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);++i)
#define reps(i,n) for(int i=1;i<=(n);++i)
#define repr(i,n) for(int i=(n);i>=0;--i)
#define cinv(n,vec) VL vec(n);rep(i,n)cin>>vec[i]
#define int long long
#define itn int
#define pb push_back
#define eb emplace_back
#define m_p make_pair
#define m_t make_tuple
#define p_p(a,b) pb(m_p(a,b))
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define RSORT(a) sort(all(a),greater<>())
#define UNIQUE(a) a.erase(unique(all(a)),a.end())
#define SZ(a) (signed)(a.size())
#define debg(a) cout<<#a<<" "<<a<<endl;
#define debgp(a) cout<<#a<<" "<<a.fi<<" : "<<a.se<<endl
#define call(a) for(auto i:a)cout<<i<<" ";cout<<endl
#define callp(a) for(auto [k,v]:a)cout<<k<<":"<<v<<endl
#define show(a) for(cont &y:a){for(cont &x:y){cout<<x<<" ";}cout<<endl;}
#define out(a) cout<<(a)<<endl
#define ENDL cout<<endl
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define fi first
#define se second
#define V vector
#define P pair
#define T tuple
#define PQ priority_queue
#define cont const auto
#define VV(type,name,y,x,a) vector<vector<type>> name = vector<vector<type>>(y,vector<type>(x,a))
typedef long long ll;
typedef double D;
typedef long double LD;
typedef string str;
typedef vector<ll> VL;
typedef pair<ll,ll> PL;
typedef vector<pair<ll,ll>> VP;
typedef tuple<ll,ll,ll> T3;
typedef tuple<ll,ll,ll,ll> T4;
typedef struct{ll to;ll cost;} Graphs;
typedef V<V<Graphs>> AdjList;
const signed INF = (1LL<<30);
const long long MOD = (1e9+7);
const long long LINF = (1LL<<60);
const long long LMAX = LLONG_MAX;
void YN(bool b){cout<<(b?"Yes":"No")<<endl;}
template<class T>inline T gcd(T a,T b){if(b==0)return a; return(gcd(b,a%b));}
template<class T>inline T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T>inline bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}

void comb(vector<vector <long long> > &v){
  for(int i = 0;i <v.size(); i++){
    v[i][0]=1;
    v[i][i]=1;
  }
  for(int k = 1;k <v.size();k++){
    for(int j = 1;j<k;j++){
      v[k][j]=(v[k-1][j-1]+v[k-1][j]);
    }
  }
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(10);

  int n,a,b;
  cin>>n>>a>>b;
  cinv(n,v);
  RSORT(v);

  D maxAve=0.0;
  rep(i,a)maxAve+=v[i];
  maxAve/=a;
  out(maxAve);

  map<ll,ll> cnt;
  for(auto e:v)cnt[e]++;
  if(cnt[v[0]]>a){
    ll sum=0;
    V<VL> nCr(60,VL(60,0));
    comb(nCr);
    for(int i=a;i<=min(cnt[v[0]],b);++i){
      sum+=nCr[n][i];
    }
    out(sum);
  }
  else{
    out(cnt[v[a-1]]);
  } 

  return 0;
}