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;

typedef long long int lli;
typedef unsigned long long int ulli;
typedef pair<int,int> pii;
typedef priority_queue< int, vector< int >, greater< int > >  minHeap;

#define ones(x) __builtin_popcount(x)
#define onesl(x) __builtin_popcountl(x)
#define onesll(x) __builtin_popcountll(x)

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define scn(n) scanf("%d",&n)
#define scnll(n) scanf("%lld",&n)
#define scn2(n,m) scanf("%d%d",&n,&m)
#define scn3(n,m,w) scanf("%d%d%d",&n,&m,&w)
#define scn2ll(n,m) scanf("%lld%lld",&n,&m)
#define atoz(v) v.begin(),v.end()
#define ratoz(v) v.rbegin(),v.rend()
#define Fill(a,v) memset(a,v,sizeof(a))
#define sz(v) v.size()
#define fi first
#define se second
#define inf 1e9
#define pi acos(-1.0)
#define sqr(x) x*x
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define ten(n) (int)1e##n
#define tenll(n) (lli)1e##n
#define mod 1000000007
#define FastIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

int Set(int N,int pos)
{
    return N=N | (1<<pos);
}
int reset(int N,int pos)
{
    return N = N & ~(1<<pos);
}
bool check(int N,int pos)
{
    return (bool)(N & (1<<pos));
}


inline int addmod(int x ,int y){return (x%mod + y%mod)%mod;}
inline int submod(int x ,int y){return (x%mod - y%mod + mod)%mod;}
inline int mulmod(int x ,int y){return (x%mod *1LL* y%mod)%mod;}
inline int nextSubMask(int i , int mask) {return (i-1)&mask;} /// returns next smallest submask

void we_r_done(){cout<<"NO"; exit(0);}

void debug(bool ok , string mssg)
{
    if(ok)return;
    cout<<mssg;
    exit(0);
}

lli dp[66][33];

lli nCr(int n, int r)
{
 if(n==r) return dp[n][r] = 1;
 if(r==0) return dp[n][r] = 1;
 if(r==1) return dp[n][r] = n;
 if(dp[n][r]) return dp[n][r];
 return dp[n][r] = nCr(n-1,r) + nCr(n-1,r-1);
}

int main()
{
    ///freopen("output.txt","w",stdout);
    ///freopen("input.txt","r",stdin);

    //FastIO;

    cout<<setprecision(10)<<fixed;



    int n, A , B;
    cin>>n>>A>>B;
    double a[n];
    map<lli,int>cnt;
    for(int i=0; i<n; i++){
        cin>>a[i];
        cnt[a[i]]++;
    }

    sort(a,a+n);

    long double ans = 0 , tot = 0;
    lli ways = 1;

    map<lli,int>taken;


    for(int i=n-1; i>=0 && (A||B); i--){

        if(a[i]<ans && A==0) break;

        tot += a[i];
        taken[a[i]]++;
        A--;
        B--;
        ans = tot/(n-i);
    }

    for(auto it : taken){
        ways *= nCr(cnt[it.fi] , it.se);
    }

    cout<<ans<<endl<<ways;

    return 0;

}

///sin and cos expect input in radians not degrees. so use , sin(degrees * pi / 180)
///using bs = bitset<MX>; // how many distinct number can be form?