★【16.6.2】Codeforces Round #355 (Div. 2) C. Vanya and Label

C. 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

z

output

3

input

V_V

output

9

input

Codeforces

output

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：给定64进制转化方案  一个字符串 问你能用多少对相同长度的字符串与给定的字符串&结果不变

有0就多了三种情况，化为二进制按位判断0的个数  对3幂运算

代码：

#include <iostream>#include <cstdio>#include <cmath>#include <map>#include <algorithm>#include <cstring>using namespace std;int poi(char ch){    if(ch>='0'&&ch<='9')return (int)ch-48;    if(ch>='A'&&ch<='Z')return (int)ch-55;    if(ch>='a'&&ch<='z')return (int)ch-61;    if(ch=='-')return 62;    if(ch=='_')return 63;}string s;int main(){    while(cin>>s)    {        long long ans=1;        for(int i=0;i<s.size();i++)        {             int cl=poi(s[i]);             for(int j=0;j<6;j++)             {                  if(!((cl>>j)&1))ans=ans*3%1000000007;             }        }        cout<<ans<<endl;    }    return 0;}