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:
"""
Author    : co_devil Chirag Garg
Institute : JIIT
"""

from __future__ import division, print_function
import itertools, os, sys, threading
from collections import deque, Counter, OrderedDict, defaultdict
import heapq
from math import ceil,floor,log,sqrt,factorial,pow,pi
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
"""from io import BytesIO, IOBase
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
else:
    from builtins import str as __str__
    str = lambda x=b'': x if type(x) is bytes else __str__(x).encode()
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._buffer = BytesIO()
        self._fd = file.fileno()
        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):
        return self._buffer.read() if self._buffer.tell() else os.read(self._fd, os.fstat(self._fd).st_size)

    def readline(self):
        while self.newlines == 0:
            b, ptr = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)), self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
            self.newlines += b.count(b'\n') + (not b)
        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)
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b'\r\n')

def print(*args, **kwargs):
    sep, file = kwargs.pop('sep', b' '), kwargs.pop('file', sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop('end', b'\n'))
    if kwargs.pop('flush', False):
        file.flush()

"""


def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int,input().split())
def li(): return list(mi())


abc = 'abcdefghijklmnopqrstuvwxyz'
abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}
mod = 1000000007
dx, dy = [-1, 1, 0, 0], [0, 0, 1, -1]
def getKey(item): return item[0]
def sort2(l): return sorted(l, key=getKey)
def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
def decimalToBinary(n): return bin(n).replace("0b", "")
def ntl(n): return [int(i) for i in str(n)]
def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
# For getting input from input.txt file
# sys.stdin = open('input.txt', 'r')

# Printing the Output to output.txt file
# sys.stdout = open('output.txt', 'w')

graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000
def dfs(v, c):
    if visited[v]:
        if col[v] != c:
            print('-1')
            exit()
        return
    col[v] = c
    visited[v] = 1
    for i in graph[v]:
        dfs(i, c ^ 1)

def bfs(d,v):
    q=[]
    q.append(v)
    visited[v]=1
    while len(q)!=0:
        x=q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i]!=1:
                visited[i]=1
                q.append(i)
        print(x)
    print(l)

def make_graph(e):
    d={}
    for i in range(e):
        x,y=mi()
        if x not in d.keys():
            d[x]=[y]
        else:
            d[x].append(y)
        if y not in d.keys():
            d[y] = [x]
        else:
            d[y].append(x)
    return d
def gr2(n):
    d={}
    for i in range(n):
        x,y=mi()
        if x not in d.keys():
            d[x]=[y]
        else:
            d[x].append(y)
    return d

def connected_components(graph):
    seen = set()
    def dfs(v):
        vs = set([v])
        component=[]
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component
    ans=[]
    for v in graph:
        if v not in seen:
            d=dfs(v)
            ans.append(d)
    return ans

s=si()
t=si()
cs=0
for i in range(len(s)):
    if s[i]=='+':
        cs+=1
    else:
        cs-=1
u=0
ct=0
for i in range(len(t)):
    if t[i]=='+':
        ct+=1
    elif t[i]=='-':
        ct-=1
    else:
        u+=1
if u==0:
    if cs==ct:
        print(1)
    else:
        print(0)
elif u>0:
    x=abs(ct-cs)
    count=0
    for i in range(int(pow(2,u))):
        y=len(decimalToBinary(i))
        z='0'*(u-y)+decimalToBinary(i)
        sum=0
        for j in range(len(z)):
            if z[j]=='0':
                sum-=1
            else:
                sum+=1
        if sum==x:
            count+=1
    print(count/pow(2,u))