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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
stdin = sys.stdin

def li(): return [int(x) for x in stdin.readline().split()]
def li_(): return [int(x)-1 for x in stdin.readline().split()]
def lf(): return [float(x) for x in stdin.readline().split()]
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(ns())
def nf(): return float(ns())

# nの逆元のリスト
def inv_mod(n:int, mod:int) -> list:
    inv = [0,1]
    for i in range(2,n+1):
        inv.append(mod - ((mod//i)*inv[mod%i]) % mod)
    return inv
       
# nの階乗のリスト
def fact(n:int, mod:int) -> list:
    fac = [1,1]
    res = 1
    for i in range(2,n+1):
        res = res*i%mod
        fac.append(res)
    return fac

# nの階乗の逆元のリスト
def fact_inv(n:int, inv:list, mod:int) -> list:
    facInv = [1,1]
    for i in range(2,n+1):
        facInv.append(facInv[i-1]*inv[i] % mod)
    return facInv
        
# 二項係数
def nCr(n:int, r:int, mod:int, fac:list, facInv:list) -> int:
    if not (0<=r and r<=n):
        return 0
    
    return ((fac[n]*facInv[r]) % mod) * facInv[n-r] % mod

from itertools import accumulate
from collections import Counter

n,a,b = li()
v = li()

v.sort(reverse=True)
v_cum = list(accumulate(v))

ans_cand = []

# 貪欲に最大値を選ぶ
for i in range(a,b+1):
    ans_cand.append(v_cum[i-1] / i)
    
ans = max(ans_cand)
opt_i = a + ans_cand.index(ans) - 1

cnt_v = Counter(v)
opt_set = v[:opt_i+1]

# 貪欲最大が何通りできるか考える
MOD = 46116646144580591
inv = inv_mod(n, MOD)
fac = fact(n, MOD)
facInv = fact_inv(n,inv, MOD)

pat = 0

if v[0] == v[opt_i]:
    for i in range(1, cnt_v[v[0]] + 1):
        pat += nCr(cnt_v[v[0]], i, MOD, fac, facInv)

else:
    cnt_opt = Counter(opt_set)
    pat = nCr(cnt_v[v[opt_i]], cnt_opt[v[opt_i]], MOD, fac, facInv)


        
print(ans)
print(pat)