Problem: Maxim has opened his own restaurant! The restaurant has got a huge table, the table's length is p meters.

Maxim has got a dinner party tonight, n guests will come to him. Let's index the guests of Maxim's restaurant from 1 to n. Maxim knows the sizes of all guests that are going to come to him. The i-th guest's size (ai) represents the number of meters the guest is going to take up if he sits at the restaurant table.

Long before the dinner, the guests line up in a queue in front of the restaurant in some order. Then Maxim lets the guests in, one by one. Maxim stops letting the guests in when there is no place at the restaurant table for another guest in the queue. There is no place at the restaurant table for another guest in the queue, if the sum of sizes of all guests in the restaurant plus the size of this guest from the queue is larger than p. In this case, not to offend the guest who has no place at the table, Maxim doesn't let any other guest in the restaurant, even if one of the following guests in the queue would have fit in at the table.

Maxim is now wondering, what is the average number of visitors who have come to the restaurant for all possible n! orders of guests in the queue. Help Maxim, calculate this number.

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of guests in the restaurant. The next line contains integers a1, a2, ..., an (1 ≤ ai ≤ 50) — the guests' sizes in meters. The third line contains integer p (1 ≤ p ≤ 50) — the table's length in meters. 

The numbers in the lines are separated by single spaces.

Output

In a single line print a real number — the answer to the problem. The answer will be considered correct, if the absolute or relative error doesn't exceed 10 - 4.

Examples

Input

3
1 2 3
3


Output

1.3333333333

Note

In the first sample the people will come in the following orders: 

  * (1, 2, 3) — there will be two people in the restaurant; 
  * (1, 3, 2) — there will be one person in the restaurant; 
  * (2, 1, 3) — there will be two people in the restaurant; 
  * (2, 3, 1) — there will be one person in the restaurant; 
  * (3, 1, 2) — there will be one person in the restaurant; 
  * (3, 2, 1) — there will be one person in the restaurant. 



In total we get (2 + 1 + 2 + 1 + 1 + 1) / 6 = 8 / 6 = 1.(3).

Solution (in PYTHON):

Solution:
def main():
    n = int(input())
    A = [int(x) for x in input().split()]
    p = int(input())

    if sum(A)<=p:
        print n
        sys.exit()

    fac = [1.0]
    for i in range(1,100):
        fac.append(fac[-1]*i)
    
    def nck(n,k):
        return fac[n]/fac[k]/fac[n-k]

    ways = 0
    for fat in range(n):
        b = A[fat]

        DP = [[0.0]*102 for _ in range(n+1)]
        DP[0][0] = 1.0
        for i in range(n):
            if i == fat:continue
            a = A[i]
            for j in reversed(range(n)):
                DPj1 = DP[j+1]
                DPj = DP[j]
                for k in range(p+1):
                    DPj1[k+a] += DPj[k]
        extra = 0
        for i in range(1,n+1):
            extra += sum(DP[i][max(p+1-b,0):p+1]) * i * fac[i] * fac[n-i-1]
        ways += extra
    print ways/fac[n]




######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

if __name__== "__main__":
  main()