暑假练习赛 006 E Vanya and Label（数学）

Vanya and LabelCrawling in process... Crawling failed Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice _ uDebug

Description

Input

Output

Sample Input

Sample Output

Hint

Description

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.

Input

The only line of the input contains a single word s (1 ≤ |s| ≤ 100 000), consisting of digits, lowercase and uppercase English letters, characters '-' and '_'.

Output

Print a single integer — the number of possible pairs of words, such that their bitwise AND is equal to string s modulo 10⁹ + 7.

Sample Input

Input

z

Output

3

Input

V_V

Output

9

Input

Codeforces

Output

130653412

Sample Output

Hint

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 ;
   }