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.

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 '_'.

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

题意:给你一个字符串s，把字符映射为一些数字，问有多少对和这个字符串长度相同的字符串逐位AND得到仍能得到字符串s

思路:看每位有多少种可能，再把他们相乘并取模，就是答案，注意到最多只有64个字符，所以O(n*n)暴力，4096*1e5大概在4e8可以在一秒内跑完

 #include<bits/stdc++.h>

 using namespace std;

 const int amn=1e5+,mod=1e9+;

 char a[amn];

 int main(){

     ios::sync_with_stdio();

     cin>>a;

     long long ans=,sum;

     int len=strlen(a),in,jg;

     for(int i=;i<len;i++){

         if(a[i]>=''&&a[i]<='')in=a[i]-'';

         else if(a[i]>='A'&&a[i]<='Z')in=a[i]-'A'+;

         else if(a[i]>='a'&&a[i]<='z')in=a[i]-'a'+;

         else if(a[i]=='-')in=;

         else in=;

         sum=;

         for(int j=;j<=;j++){

             for(int k=;k<=;k++){

                 jg=j&k;

                 if(jg==in)sum++;

             }

         }

         ans=((ans%mod)*(sum%mod))%mod;

     }

     printf("%lld\n",ans);

 }

 /***

 给你一个字符串s，把字符映射为一些数字，问有多少对和这个字符串长度相同的字符串逐位AND得到仍能得到字符串s

 看每位有多少种可能，再把他们相乘并取模，就是答案，注意到最多只有64个字符，所以O(n*n)暴力，4096*1e5大概在4e8可以在一秒内跑完

 ***/