Codeforces Round #355 (Div. 2) C

Description

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  

给我们一下字符串，问我们可以通过&后，值依然和题目上规定的相同

我们当然可以先把0~63的先全部&一下，存下结果，然后，就一个字符一个字符的判断就好啦~

#include<iostream>
#include<stdio.h>
#include<map>
using namespace std;
map<int,int> q;
int main()
{
    for(int i=0;i<=63;i++)
    {
        for(int j=0;j<=63;j++)
        {
            q[i&j]++;
        }
    }
    int pos;
    long long num=1;
    string s;
    cin>>s;
    for(int i=0;i<s.length();i++)
    {
        if(s[i]>='0'&&s[i]<='9')
        {
            pos=s[i]-'0';
        }
        else if(s[i]>='A'&&s[i]<='Z')
        {
            pos=s[i]-'A'+10;
        }
        else if(s[i]>='a'&&s[i]<='z')
        {
            pos=s[i]-'a'+36;
        }
        else if(s[i]=='-')
        {
            pos=62;
        }
        else if(s[i]=='_')
        {
            pos=63;
        }
        num=num*q[pos]%1000000007;
    }
    cout<<num<<endl;
    return 0;
}