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 JAVA):

Solution:


/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */





//import com.sun.xml.internal.ws.util.StringUtils;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
import java.util.stream.Collectors;





/**
 *
 * @author george
 */

public class main {
    /*
public static boolean isPrime(long n) {
    if(n < 2) return false;
    if(n == 2 || n == 3) return true;
    if(n%2 == 0 || n%3 == 0) return false;
    long sqrtN = (long)Math.sqrt(n)+1;
    for(long i = 6L; i <= sqrtN; i += 6) {
        if(n%(i-1) == 0 || n%(i+1) == 0) return false;
    }
    return true;
}*/
/*
    private static long f(long l) {
        throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
    }

public static boolean contains(final int[] arr, final int key) {
    return Arrays.stream(arr).anyMatch(i -> i == key);
}
 static boolean isSubSequence(String str1, String str2, int m, int n)
    {
       
        if (m == 0) 
            return true;
        if (n == 0) 
            return false;
             
        
        if (str1.charAt(m-1) == str2.charAt(n-1))
            return isSubSequence(str1, str2, m-1, n-1);
 
        
        return isSubSequence(str1, str2, m, n-1);
    }

    static int gcdThing(int a, int b) {
    BigInteger b1 = BigInteger.valueOf(a);
    BigInteger b2 = BigInteger.valueOf(b);
    BigInteger gcd = b1.gcd(b2);
    return gcd.intValue();
}
    public static boolean checkAnagram(String str1, String str2) {
        int i=0;
        for (char c : str1.toCharArray()) {
        i = str2.indexOf(c, i) + 1;
        if (i <= 0) { return false; }
    }
    return true;

}
    
    public static Set primeFactors(int num)
    {
        Set<Integer>Prime=new HashSet<Integer>();
        // Print the number of 2s that divide n
        int tw=2;
        while (num%2==0)
        {
            Prime.add(tw);
            num /= 2;
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(num); i+= 2)
        {
            // While i divides n, print i and divide n
            while (num%i == 0)
            {
                Prime.add(i);
                num/= i;
            }
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (num> 2)
            Prime.add(num);        
        return Prime;
    }
    public static String Winner(int [] arr)
    {
        int maxsum=-1;
        int temp=0;
        List<Integer> L=Arrays.stream(arr).boxed().collect(Collectors.toList());
        int count=0;
        String winner="First";
        while(L.isEmpty()!=true){
           if(count%2==0){winner="First";
           //take max odd interval
            for (int i = 0; i < L.size(); i++) {
                
            }
            }
           else{winner="Second";
           
           
           
           
           
           
           
           }
           count++;
        }
    
    
    return winner;
    }
    public static String changeCharInPosition(int position, char ch, String str){
    char[] charArray = str.toCharArray();
    charArray[position] = ch;
    return new String(charArray);
    }
    public static boolean CheckCanObtainYfromX(String y,String x){
        boolean cond=true;
        String help="";
        for (int i = 0; i < y.length(); i++) {
            help="";help+=y.charAt(i);
            if(y.contains(help)==false){x=x.replaceFirst(help,"A");cond=false;break;}
            
        }
        return cond;
    }
   public static int LongestPalindrome(String x){
    int n=x.length();
    boolean table[][] = new boolean[n][n];
    int maxLength = 1;
     for (int i = 0; i < n; ++i)
            table[i][i] = true;
   int start = 0;
    for (int i = 0; i < n - 1; ++i) {
            if (x.charAt(i) == x.charAt(i + 1)) {
                table[i][i + 1] = true;
                start = i;
                maxLength = 2;
            }
        }
     for (int k = 3; k <= n; ++k) {
            for (int i = 0; i < n - k + 1; ++i) 
            {
                int j = i + k - 1;
               if (table[i + 1][j - 1] && x.charAt(i) == x.charAt(j)) {
                    table[i][j] = true;
                if (k > maxLength) {
                        start = i;
                        maxLength = k;
                    }
                }
            }
     }
   return maxLength;
   }
   public static long choose(long total, long choose){
    if(total < choose)
        return 0;
    if(choose == 0 || choose == total)
        return 1;
    return choose(total-1,choose-1)+choose(total-1,choose);
}
    public static long bitwiseAdd(int n1, int n2) {
        int x = n1, y = n2;
        int xor, and, temp;
        and = x & y;
        xor = x ^ y;

        while (and != 0) {
            and <<= 1;
            temp = xor ^ and;
            and &= xor;
            xor = temp;
        }
        return (xor%2);
    }
    public static boolean compare(String a,String b){
        for (int i = 0; i < a.length(); i++) {
            if(a.charAt(i)!=b.charAt(i)&&a.charAt(i)!='?'){return false;}
        }
  return true;
    }*/
/*public static boolean isUniqueChars(String str) {
    // short circuit - supposed to imply that
    // there are no more than 256 different characters.
    // this is broken, because in Java, char's are Unicode,
    // and 2-byte values so there are 32768 values
    // (or so - technically not all 32768 are valid chars)
    if (str.length() > 256) {
        return false;
    }
    // checker is used as a bitmap to indicate which characters
    // have been seen already
    int checker = 0;
    for (int i = 0; i < str.length(); i++) {
        // set val to be the difference between the char at i and 'a'
        // unicode 'a' is 97
        // if you have an upper-case letter e.g. 'A' you will get a
        // negative 'val' which is illegal
        int val = str.charAt(i) - 'a';
        // if this lowercase letter has been seen before, then
        // the corresponding bit in checker will have been set and
        // we can exit immediately.
        if ((checker & (1 << val)) > 0) return false;
        // set the bit to indicate we have now seen the letter.
        checker |= (1 << val);
    }
    // none of the characters has been seen more than once.
    return true;
}*/
 private static int [] PossibilitiesAsBinaryTruthTable(int n) {
        int rows = (int) Math.pow(2,n);
        int [] possibleTrack=new int[rows];
        int trackPerRow;
        for (int i=0; i<rows; i++) {
            trackPerRow=0;
            for (int j=n-1; j>=0; j--) {
                if(((i/(int) Math.pow(2, j))%2)==0){trackPerRow--;}
                else if(((i/(int) Math.pow(2, j))%2)==1){trackPerRow++;}
            }
            possibleTrack[i]=trackPerRow;
        }
        return possibleTrack;
    }
    public static void main (String []  args) throws IOException 
    {
        Scanner s=new Scanner(System.in);
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));
        String Destination=br.readLine();
        String DefectedTrack=br.readLine();
        int pos=0;
        for (int i = 0; i < Destination.length(); i++) {
            if(Destination.charAt(i)=='+'){pos++;}
            else{pos--;}
        }
        int Defectedpos=0;
        int unreconized=0;
        for (int i = 0; i < DefectedTrack.length(); i++) {
            if(DefectedTrack.charAt(i)=='+'){Defectedpos++;}
            else if(DefectedTrack.charAt(i)=='-'){Defectedpos--;}
            else{unreconized++;}
        }
       if(!DefectedTrack.contains("?")&&Defectedpos==pos){System.out.print((double)1/1);System.exit(0);}
       //now we have total of 2^unreconized posibilities and we have to check for all of them
       //to calculate the probability we need to caculate the number of valid ways and then devide them over the total 
       int [] possibilities=PossibilitiesAsBinaryTruthTable(unreconized);
       int ValidPossibilities=0;
        for (int i = 0; i < possibilities.length; i++) {
            if(Defectedpos+possibilities[i]==pos){ValidPossibilities++;}
        }
       System.out.print(ValidPossibilities/(Math.pow(2, unreconized)));
       
    }  
   }