Dreamoon and WiFi



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.

Sample test(s)

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.

n个字符，然后接受到的样子也是n个，然后判断多少概率是保证结果没有错的。即字符个数相等就可以。其实就是一个数学题。首先看原来的+和-个数，再算后来的+-？个数，因为只有2种字符，所以一个确定时，另一个也就确定了。即m中取n的排列方式，然后写一个函数来表示一下。最后用符合的例子除以总例子，就是概率了。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
	char a[12], b[12];
	int i, j, m, n, pre1, pre2, temp1, temp2, temp3, len1, len2, temp, c1, c2, pre;
	double ans;
	scanf("%s", a); getchar();
	scanf("%s", b);
	len1 = strlen(a); len2 = strlen(b);
	pre1 = pre2 = temp1 = temp2 = temp3 = 0;
	for (i = 0; i < len1; i++)
	{
		if (a[i] == '+')
			pre1++;
		else
			pre2++;
	}
	for (i = 0; i < len2; i++)
	{
		if (b[i] == '+')
			temp1++;
		else if (b[i] == '-')
			temp2++;
		else temp3++;
	}

	if (temp1 == pre1&&temp2 == pre2)
		cout << "1.000000000000\n";
	else if (temp3 == 0)
		cout << "0.000000000000\n";
	else if (temp1 + temp3 < pre1)
		cout << "0.000000000000\n";
	else if (temp2 + temp3 < pre2)
		cout << "0.000000000000\n";
	else
	{
		temp1 = pre1 - temp1;
		temp2 = pre2 - temp2;
		if (temp1 == 0 || temp2 == 0)
			temp = 1;
		else
		{
			c1 = 1; c2 = 1;
			for (i = temp3; i > temp3 - temp1; i--)
				c1 = c1*i;
			for (i = 1; i <= temp1; i++)
				c2 = c2*i;
			temp = c1 / c2;
		}
		pre = 1;
		for (i = 1; i <= temp3; i++)
			pre = pre * 2;
		ans = temp*1.0 / pre;
		printf("%.12lf\n", ans);
	}
	return 0;
}