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:
import java.util.*;
import java.lang.*;
import java.io.*;

// SHIVAM GUPTA :
//NSIT
//decoder_1671
//BEING PERFECTIONIST IS NOT AN OPTION



// ASCII = 48 + i ;

// SHIVAM GUPTA :


/* Name of the class has to be "Main" only if the class is public. */



public class Main
{
     
     
     
    static PrintWriter out;
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		String next(){
			while(st==null || !st.hasMoreElements()){
				try{
					st= new StringTokenizer(br.readLine());
				}
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		String nextLine(){
			String str = "";
			try{
				str=br.readLine();
			}
			catch(IOException e){
				e.printStackTrace();
			}
			return str;
		}
	}
	





/////////////////////////////////////////////////////////////////////////////////////

public static int  min(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[0];
       
}
 
 /////////////////////////////////////////////////////////////////////////////
 
 
public static int  max(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[3];
       
}
 
/////////////////////////////////////////////////////////////////////////////////// 
static int sieve =  1000000 ;

 
static boolean[] prime =  new boolean[sieve + 1] ;

 
public static void sieveOfEratosthenes() 
    { 
        // FALSE == prime
        
        // TRUE ==  COMPOSITE
        
        // FALSE== 1
       
        for(int i=0;i< sieve + 1;i++) 
            prime[i] = false; 
          
        for(int p = 2; p*p <= sieve; p++) 
        { 
           
            if(prime[p] == false) 
            { 
                
                for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = true; 
            } 
        } 
          
       
       
    
    } 
 
///////////////////////////////////////////////////////////////////////////////////


public static String reverse(String input)
{
    String op = "" ;
    
    for(int i = 0; i < input.length() ; i++ )
    {
        op = input.charAt(i)+  op ; 
    }
    
return op ;
}


///////////////////////////////////////////////////////////////////////////////////////


public static void sortD(int[] arr)
{
      sort(arr ,0 , arr.length-1) ;
      
      int i =0 ; int j = arr.length -1  ;
      
      while( i < j)
      {
            int temp = arr[i] ;
            arr[i] =arr[j] ;
            arr[j] = temp ;
            i++ ; j-- ;
      }
      
      
      
      return ;
}

///////////////////////////////////////////////////////////////////////////////////////

public static boolean sameParity(long a  ,long b )
{
      
      if(a%2 == b %2)
      {
            return true  ;
      }
      else{
            return false  ;
      }
}


///////////////////////////////////////////////////////////////////////////////////////

public static boolean sameParity(int a  ,int b )
{
      
      if(a%2 == b %2)
      {
            return true  ;
      }
      else{
            return false  ;
      }
}

////////////////////////////////////////////////////////////////////////////////////

public static boolean isPossibleTriangle(int a  ,int b , int c)
{
      if( a + b > c && c+b > a && a +c > b)
      {
            return true  ;
      }
      
      return false  ;
      
      
}

/////////////////////////////////////////////////////////////////////
    
public static int gcd(int a, int b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 


}

////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c , int d )
{

int temp = lcm(a,b , c) ;


 
 int ans = lcm(temp ,d ) ;

return ans  ;


}

//////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c )
{

int temp = lcm(a,b) ;

int ans =  lcm(temp ,c) ;

return ans  ;


}

////////////////////////////////////////////////////////////////////////////////////////
    
public static int lcm(int a , int b )
{

int gc = gcd(a,b);

return (a*b)/gc ;




}

/////////////////////////////////////////////////////////////////////////////////////////////

static boolean isPrime(int n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(int i = 2; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
      
      
}
//////////////////////////////////////////////////////////////////////////////////////////
public static long nCr(int n,int k)
{
    long ans=1;
    k=k>n-k?n-k:k;
    int j=1;
    for(;j<=k;j++,n--)
    {
        if(n%j==0)
        {
            ans*=n/j;
        }else
        if(ans%j==0)
        {
            ans=ans/j*n;
        }else
        {
            ans=(ans*n)/j;
        }
    }
    return ans;
}
///////////////////////////////////////////////////////////////////////////////////////////

public static ArrayList<Integer> allFactors(int n)
{   
      ArrayList<Integer> list = new ArrayList<>() ;
      
    for(int i = 1; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)
              {
                    list.add(i) ;
              }
              else{
                    list.add(i) ;
                    list.add(n/i) ;
                    
              }
          }
    }
      
     return list ; 
      
      
}



/////////////////////////////////////////////////////////////////////////////

public static boolean isPowerOfTwo(int n) 
{ 
    if(n==0) 
    return false; 
  
if(((n ) & (n-1)) == 0 ) return true ;
else return false  ;

} 
   
////////////////////////////////////////////////////////////////////////////////////   
   
    public static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 
    
 ///////////////////////////////////////////////////////////////////////////////////////// 
  
   static final int MAXN = 100001; 
       
    
    static int spf[] = new int[MAXN]; 
   
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i=2; i<MAXN; i++) 
       
           
            spf[i] = i; 
       
       
        for (int i=4; i<MAXN; i+=2) 
            spf[i] = 2; 
       
        for (int i=3; i*i<MAXN; i++) 
        { 
           
            if (spf[i] == i) 
            { 
               
                for (int j=i*i; j<MAXN; j+=i) 
       
                    if (spf[j]==j) 
                        spf[j] = i; 
            } 
        } 
    } 
       
// The above code works well for n upto the order of 10^7.
// Beyond this we will face memory issues.

// Time Complexity: The precomputation for smallest prime factor is done in O(n log log n)
// using sieve.
// Where as in the calculation step we are dividing the number every time by 
// the smallest prime number till it becomes 1.
// So, let’s consider a worst case in which every time the SPF is 2 .
// Therefore will have log n division steps.


// Hence, We can say that our Time Complexity will be O(log n) in worst case.


    static Vector<Integer> getFactorization(int x) 
    { 
        Vector<Integer> ret = new Vector<>(); 
        while (x != 1) 
        { 
            ret.add(spf[x]); 
            x = x / spf[x]; 
        } 
        return ret; 
    } 
       
  
   
public static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
  public static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

 /////////////////////////////////////////////////////////////////////////////////////////
 
	public static int knapsack(int[] weight,int value[],int maxWeight){

        
        int n=  value.length ;
        
	
	//dp[i] stores the profit with KnapSack capacity "i" 
    int []dp = new int[maxWeight+1]; 
      
    //initially profit with 0 to W KnapSack capacity is 0 
    Arrays.fill(dp, 0); 
  
    // iterate through all items 
    for(int i=0; i < n; i++)  
      
        //traverse dp array from right to left 
        for(int j = maxWeight; j >= weight[i]; j--) 
            dp[j] = Math.max(dp[j] , value[i] + dp[j - weight[i]]); 
              
    /*above line finds out maximum of dp[j](excluding ith element value) 
    and val[i] + dp[j-wt[i]] (including ith element value and the 
    profit with "KnapSack capacity - ith element weight") */
    return dp[maxWeight]; 
	}

///////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////


// to return max sum of any subarray in given array
public static long kadanesAlgorithm(long[] arr)
{
      long[] dp = new long[arr.length] ;
      
      dp[0] = arr[0] ;
      long max =  dp[0] ;
      
      
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0)
            {
                  dp[i] = dp[i-1] + arr[i] ;
            }
            else{
                  dp[i] = arr[i] ;
            }
            
            if(dp[i] >  max)max = dp[i] ;
            
      }
      
      return max  ;
      
}
/////////////////////////////////////////////////////////////////////////////////////////////
public static long kadanesAlgorithm(int[] arr)
{
      long[] dp = new long[arr.length] ;
      
      dp[0] = arr[0] ;
      long max =  dp[0] ;
      
      
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0)
            {
                  dp[i] = dp[i-1] + arr[i] ;
            }
            else{
                  dp[i] = arr[i] ;
            }
            
            if(dp[i] >  max)max = dp[i] ;
            
      }
      
      return max  ;
      
}

      
///////////////////////////////////////////////////////////////////////////////////////         

public static void main (String[] args) throws java.lang.Exception
{
  FastReader scn = new FastReader() ;

String s1 = scn.next() ;
String s2 = scn.next() ;

int s1p = 0; int s1m = 0 ;
int s2p = 0; int s2m = 0 ;  int s2q = 0 ;

int n = s1.length() ;

for(int i = 0; i < n  ; i++)
{
      char ch  = s1.charAt(i);
      if(ch == '+')s1p++ ;
      else s1m++ ;
}

for(int i = 0; i < n  ; i++)
{
      char ch  = s2.charAt(i);
      if(ch == '+')s2p++ ;
      else if(ch == '-') s2m++ ;
      else s2q++ ;
      
}


if(s2q == 0)
{
      if(s1p == s2p)
      {
            out.println(1) ;
            
      }
      else{
            out.println(0) ;
      }
}

else{
      
      int x  = s1p- s1m ;
      int y =  s2p - s2m ;
      
      int z  = y-x ;
      
      if(z<0)z= -1*z ;
      
      if(sameParity(s2q,z) == false )out.println(0) ;
      
      else{
            
         if((int)z > s2q )
         {
               out.println(0) ;
         }
         else if((int)z == s2q)
         {
               double p = Math.pow(2,s2q) ;
               out.println(1/p) ;
         }
         else{
              
              int cal = s2q-z  ;
              
              cal = cal/2  ;
              
            long cases = nCr(s2q , cal) ;
            
             double p = Math.pow(2,s2q) ;
               
              out.println(cases/p) ;  
         }
            
            
      }
      
      
      
      
      
}

 
out.flush() ;
      
}


}
  
 class Pair implements Comparable<Pair>
{
  int val ;
  
  String str  ;
      
Pair(int val , String str)
{
      this.val = val ;
      this.str = str  ;
}

  public int compareTo(Pair o)
    {
        return this.val - o.val;
    }
 
     
}