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 static java.lang.System.out;
import static java.lang.Math.*;

import java.io.DataInputStream; 
import java.io.FileInputStream; 
import java.io.IOException; 
import java.io.InputStreamReader;
import java.math.BigInteger; 

public class Main 
{ 

	static int mod = 1000000007;
	static int MOD = 1000000007;
	static final long M = (int)1e9+7;
	 
	static class Pair {
		int first;
		int second;

		public Pair(int first, int second) {
			this.first = first;
			this.second = second;
		}
	}
	
	static class Reader 
	{ 
		final private int BUFFER_SIZE = 1 << 16; 
		private DataInputStream din; 
		private byte[] buffer; 
		private int bufferPointer, bytesRead; 

		public Reader() 
		{ 
			din = new DataInputStream(System.in); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public Reader(String file_name) throws IOException 
		{ 
			din = new DataInputStream(new FileInputStream(file_name)); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public String next() throws IOException 
		{ 
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c; 
			while ((c = read()) != -1) 
			{ 
				if (c == '\n') 
					break; 
				buf[cnt++] = (byte) c; 
			} 
			return new String(buf, 0, cnt); 
		} 
		
		public int nextInt() throws IOException 
		{ 
			int ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do
			{ 
				ret = ret * 10 + c - '0'; 
			} while ((c = read()) >= '0' && c <= '9'); 

			if (neg) 
				return -ret; 
			return ret; 
		}
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}

		public long nextLong() throws IOException 
		{ 
			long ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 
			if (neg) 
				return -ret; 
			return ret; 
		} 

		public double nextDouble() throws IOException 
		{ 
			double ret = 0, div = 1; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 

			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 

			if (c == '.') 
			{ 
				while ((c = read()) >= '0' && c <= '9') 
				{ 
					ret += (c - '0') / (div *= 10); 
				} 
			} 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		private void fillBuffer() throws IOException 
		{ 
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
			if (bytesRead == -1) 
				buffer[0] = -1; 
		} 

		private byte read() throws IOException 
		{ 
			if (bufferPointer == bytesRead) 
				fillBuffer(); 
			return buffer[bufferPointer++]; 
		} 

		public void close() throws IOException 
		{ 
			if (din == null) 
				return; 
			din.close(); 
		}

	}

	
	public static boolean isPrime(int n) {
		if(n == 1)
		{
			return false;
		}
		for(int i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	}
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}


	public static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
	
	public static int phi(int n)  //euler totient function
    { 
        int result = 1; 
        for (int i = 2; i < n; i++) 
            if (gcd(i, n) == 1) 
                result++; 
        return result; 
    }
	
	
	public static int[] computePrefix(int arr[], int n)
	{
		int[] prefix = new int[n];
		prefix[0] = arr[0];
		for(int i = 1;i<n;i++)
		{
			prefix[i] = prefix[i-1]+arr[i];
		}
		
		return prefix;
	}
	
	
	public static int fastPow(int x, int n) //include mod at each step if asked and in args of fn too
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	

	
	
	static int LowerBound(int a[], int x) {
		  int l=-1,r=a.length;
		  while(l+1<r) {
		    int m=(l+r)>>>1;
		    if(a[m]>=x) r=m;
		    else l=m;
		  }
		  return r;
		}
	
	
	static int UpperBound(int a[], int x) {
	    int l=-1,r=a.length;
	    while(l+1<r) {
	       int m=(l+r)>>>1;
	       if(a[m]<=x) l=m;
	       else r=m;
	    }
	    return l+1;
	 }
	
	
	   static int power(int x, int y, int p)
	    {
	 
	        int res = 1;
	 
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	 
	        return res;
	    }
	 
	    static int modInverse(int n, int p)
	    {
	        return power(n, p - 2, p);
	    }
	    
	    //Fermat's little theorem for nCr % p.
	    static int nCrModPFermat(int n, int r,
	                             int p)
	    {
	 
	          if (n<r) 
	              return 0;
	        if (r == 0)
	            return 1;
	        int[] fac = new int[n + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[n] * modInverse(fac[r], p)
	                % p * modInverse(fac[n - r], p)
	                % p)
	            % p;
	    }
	    

		public static void Sort(int[] a) {
			List<Integer> l=new ArrayList<>();
			for (int i:a) l.add(i);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
	  
	public static void main(String[] args) throws IOException 
	{ 
		Reader sc = new Reader();
		
		String l1=sc.next();
		int count1=0, count2=0;
		for (char c:l1.toCharArray()) if (c=='+') count1++;
		int qs=0;
		String l2=sc.next();
		for (char c:l2.toCharArray())
			if (c=='+') count2++;
			else if (c=='?') qs++;
		int correct=0, incorrect=0;
		for (int i=0; i<1<<qs; i++) if (Integer.bitCount(i)+count2==count1) correct++; else incorrect++;
		
		System.out.printf("%.10f\n", (correct)/(double)(correct+incorrect));
		
	}
}