[暴力枚举]Codeforces Vanya and Label

Vanya and Label

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

Copy

z

output

Copy

3

input

Copy

V_V

output

Copy

9

input

Copy

Codeforces

output

Copy

130653412

Note

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

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
 ***/