While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwise AND for these two words represented as a integers in base 64 and got new word. Now Vanya thinks of some string s and wants to know the number of pairs of words of length |s| (length of s), such that their bitwise AND is equal to s. As this number can be large, output it modulo 10⁹ + 7.

To represent the string as a number in numeral system with base 64 Vanya uses the following rules:

• digits from '0' to '9' correspond to integers from 0 to 9;
• letters from 'A' to 'Z' correspond to integers from 10 to 35;
• letters from 'a' to 'z' correspond to integers from 36 to 61;
• letter '-' correspond to integer 62;
• letter '_' correspond to integer 63.

For a detailed definition of bitwise AND we recommend to take a look in the corresponding article in Wikipedia.

In the first sample, there are 3 possible solutions:

1. z&_ = 61&63 = 61 = z
2. _&z = 63&61 = 61 = z
3. z&z = 61&61 = 61 = z

   /*
   
   有几个0，就有几个三
   
   */
   
   #include <string.h>
   
   #include <iostream>
   
   #include <algorithm>
   
   #include <stdio.h>
   
   #define N 100010
   
   using namespace std;
   
   const int mod =1e9+;
   
   /*void inti()
   
   {
   
       for(int i=0;i<=63;i++)
   
       {
   
           int s=0;
   
           while(i)
   
           {
   
               if(i%2==0) s++;
   
               i/=2;
   
           }
   
           ans[i]=s;
   
       }
   
   }
   
   */
   
   int get(char c)
   
   {
   
       if(c>=''&&c<='')return c-'';
   
       if(c>='A'&&c<='Z')return c-'A'+;
   
       if(c>='a'&&c<='z')return c-'a'+;
   
       if(c=='-')return ;
   
       if(c=='_')return ;
   
   }
   
   int main()
   
   {
   
       //freopen("in.txt","r",stdin);
   
       char ch[N];
   
       long long s=;
   
       scanf("%s",&ch);
   
       for(int i=;ch[i];i++)
   
       {
   
           long long p=get(ch[i]);
   
           for(int j=;j<;j++)
   
               if(!((p>>j)&))
   
                   s=s*%mod; //只有三种
   
       }
   
       printf("%lld\n",s);
   
       return ;
   
   }