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: Accepted)

/*
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*/

#include <bits/stdc++.h>


#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)
#define int long long
#define mod (int)(1000000007)

#define intt long long
#define P pair<int,int>

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)

#define ALL(v) v.begin(),v.end()

using namespace std;

//Library
//モッドパウ
int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % mod;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}


//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}


//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}


template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}


#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",i)
#define fendl printf("\n")
//Library-End

class modInt {
    int value, modulo;
public:
    modInt() : value(0), modulo(mod) { value = 0; }
    template<typename T>
    modInt(T value = 0, int modulo = mod) : value(value), modulo(modulo) {
        if (value < 0) {
            value = -value;
            value %= modulo;
            value = -value;
            value += modulo;
        }
        this->value = value % modulo;
    }
    inline operator int()const { return value; }
    inline modInt& operator+=(modInt x) {
        value += x.value;
        if (value >= modulo)value -= modulo;
        return *this;
    }
    inline modInt& operator++() {
        if (value == modulo - 1)value = 0;
        else value++;
        return *this;
    }
    inline modInt& operator-()const {
        return modInt(0) -= *this;
    }
    inline modInt& operator-=(modInt x) {
        value -= x.value;
        if (value < 0)value += modulo;
        return *this;
    }
    inline modInt& operator--() {
        if (value == 0)value = modulo - 1;
        else value--;
        return *this;
    }
    inline modInt& operator*=(modInt x) {
        value = value * x.value % modulo;
        return *this;
    }
    inline modInt& operator/=(modInt x) {
        return operator*=(x.inv());
    }
    inline modInt inv() {
        return modpow(*this, modulo - 2);
    }
    template<typename T> modInt operator+(T x) { return modInt(*this) += x; }
    template<typename T> modInt& operator+=(T x) { return operator+=(modInt(x, modulo)); }
    template<typename T> modInt operator-(T x) { return modInt(*this) -= x; }
    template<typename T> modInt& operator-=(T x) { return operator-=(modInt(x, modulo)); }
    template<typename T> modInt operator*(T x) { return modInt(*this) *= x; }
    template<typename T> modInt& operator*=(T x) { return operator*=(modInt(x, modulo)); }
    template<typename T> modInt operator/(T x) { return modInt(*this) /= x; }
    template<typename T> modInt& operator/=(T x) { return operator/=(modInt(x, modulo)); }
};
istream& operator>>(istream& ist, modInt& x) {
    int a;
    ist >> a;
    x = a;
    return ist;
}
int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    //    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    intt kake=1, waru=1;
    for (int i=x-y+1; i<=x; i++) {
        kake*=i;
        waru*=i-(x-y+1)+1;
        int g=gcd(kake, waru);
        kake/=g;
        waru/=g;
    }
    return kake;
//    return kai(x, y) / kai(y,y);
}

double n,a,b,v[100004],hei,cnt,basho;
intt ans;
signed main() {
    cin>>n>>a>>b;
    rep(i,n)cin>>v[i];
    sort(v,v+(int)n,greater<double>());
    rep(i,a){
        hei+=v[i];
    }
    basho=-1;
    rep(i,n){
        if (v[i]==v[(int)a-1]) {
            if (basho==-1) {
                basho=i;
            }
            cnt++;
        }
    }
    hei/=a;

    if (!basho) {
        for (int i=a; i<=min(b,cnt); i++) {
            ans+=comb(cnt,i);
//            cout<<comb(cnt,i)<<endl;
        }
    }else{
        ans=comb(cnt,a-basho);
    }
    printf("%.10lf\n%lld\n",hei,ans);
}