Vanya and Label
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:
- z&_ = 61&63 = 61 = z
- _&z = 63&61 = 61 = z
- z&z = 61&61 = 61 = z
题意：
给出字符串转化为数字，问有多少个 位 & 还是原来的数字
思路：
把这些字符转化成数字后变成2进制,看有多少个0,如果有n个0,那么就是3的n次方; 因为每一个为0的位置上可以有0&1,1&0,0&0,这3种情况,最后结果就是3的n次方了,快速幂 AC代码：
# include <iostream>
using namespace std;
const int mod = 1e9+7;
string s;
int f(char ch)
{
if(ch >= ‘0‘ && ch <= ‘9‘)
return ch - ‘0‘;
else if(ch >= ‘A‘ && ch <= ‘Z‘)
return ch-‘A‘+10;
else if(ch >= ‘a‘ && ch <= ‘z‘)
return ch - ‘a‘ + 36;
else if(ch == ‘-‘)
return 62;
else if(ch == ‘_‘)
return 63;
}
int main()
{
cin >> s;
long long ans = 1;
for(int i = 0; i < s.size();i++)
{
int t = f(s[i]);
for(int j = 0; j < 6; j++)
if(!((t >> j) & 1 ))
ans = ans * 3 % mod;
}
cout << ans << endl;
}