Problem: Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.

Each command is one of the following two types: 

  1. Go 1 unit towards the positive direction, denoted as '+'
  2. Go 1 unit towards the negative direction, denoted as '-'



But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5). 

You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?

Input

The first line contains a string s1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}. 

The second line contains a string s2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.

Lengths of two strings are equal and do not exceed 10.

Output

Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9.

Examples

Input

++-+-
+-+-+


Output

1.000000000000


Input

+-+-
+-??


Output

0.500000000000


Input

+++
??-


Output

0.000000000000

Note

For the first sample, both s1 and s2 will lead Dreamoon to finish at the same position  + 1. 

For the second sample, s1 will lead Dreamoon to finish at position 0, while there are four possibilites for s2: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5. 

For the third sample, s2 could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position  + 3 is 0.

Solution (in PYTHON3):

Solution:
#------------------------------what is this I don't know....just makes my mess faster--------------------------------------
import os
import sys
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
class FastIO(IOBase):
	newlines = 0
 
	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None
 
	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines = 0
		return self.buffer.read()
 
	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b"\n") + (not b)
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines -= 1
		return self.buffer.readline()
 
	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		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")
 
#----------------------------------Real game starts here--------------------------------------

#_______________________________________________________________#
def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): #return 0 if all are greater or equal to
	answer = -1 
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): #return n-1 if all are small or equal
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub):
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans

def sieve_of_eratosthenes(n):
	ans = []
	arr = [1]*(n+1)
	arr[0],arr[1], i = 0, 0, 2
	while(i*i <= n):
		if arr[i] == 1:
			j = i+i
			while(j <= n):
				arr[j] = 0
				j += i
		i += 1
	for k in range(n):
		if arr[k] == 1:
			ans.append(k)
	return ans
def nc2(x):
	if x == 1:
		return 0
	else:
		return x*(x-1)//2
'''_______________________________________________________________
...........................................................................
.   _-----_                      |                    '-.               _/.
. _-       -_                    |                    _  |            _|  .
.|           |                ___|___    _.--.       |▒|  '.        _|    .
.|           |	  __           	 |      | ..-.|       _    |      _|      .
.|-----------| _-'  '-_   _-''-_ |      | |    __    |▒|    '-.  -'       .
.|▒▒▒▒▒▒▒▒▒▒▒||▒|      | ||      |      | | _-'  '-_ |▒|       |▒|        .
.|-----------||▒|_____-' ||_     |      | || |	  | ||▒|      _|▒|        .
.|           ||▒|     _ _   '-._ |     || || |____| ||▒|   _-'   '|_      .
.|           | -_-___-   -_____.''-__-' |_|'-______-'|▒| _|         '-_   .
........................................................................... 
___________________THIS IS AESTROIX CODE________________________
		                KARMANYA GUPTA
'''

'''
    ____________________
   ▄███████▀▀▀▀▀▀███████▄
░▐████▀▒▒Aestroix▒▒▀██████
░███▀▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▀████
░▐██▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒████▌         ________________
░▐█▌▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒████▌  ? ?   |▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒|
░░█▒▒▄▀▀▀▀▀▄▒▒▄▀▀▀▀▀▄▒▒▐███▌   ?    |___CM ONE DAY___|
░░░▐░░░▄▄░░▌▐░░░▄▄░░▌▒▐███▌     ? ? |▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒|
░▄▀▌░░░▀▀░░▌▐░░░▀▀░░▌▒▀▒█▌    ? ?    
░▌▒▀▄░░░░▄▀▒▒▀▄░░░▄▀▒▒▄▀▒▌      ? 
░▀▄▐▒▀▀▀▀▒▒▒▒▒▒▀▀▀▒▒▒▒▒▒█     ? ? 
░░░▀▌▒▄██▄▄▄▄████▄▒▒▒▒█▀       ? 
░░░░▄█████████ ████	       █▒▒▐▌
░░░▀███▀▀████▀█████▀▒▌
░░░░░▌▒▒▒▄▒▒▒▄▒▒▒▒▒▒▐
░░░░░▌▒▒▒▒▀▀▀▒▒▒▒▒▒▒▐
'''
def check(sa,sb,string):
	if len(string) == 0:
		if sa == sb:
			return 1
		else:
			return 0
	if string[0] == '+':
		return check(sa,sb+1,string[1:])
	elif string[0] == '-':
		return check(sa, sb-1, string[1:])
	else:
		return check(sa,sb+1,string[1:])+check(sa,sb-1,string[1:])
for _ in range(1):
	a = input()
	b = input()
	sa,un = 0,0
	for i in range(len(a)):
		if a[i] == '+':
			sa += 1
		else:
			sa -= 1
		if b[i] == '?':
			un += 1
	res = check(sa,0,b)
	print("%.12f"%(res/(2**un)))