CodeForce 677C - Vanya and Label

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:

1. z&_ = 61&63 = 61 = z
2. _&z = 63&61 = 61 = z
3. 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;
}