Codeforces 677C. Vanya and Label 位操作

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

 

传送门：http://codeforces.com/contest/677/problem/C

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题意：一个字符串s，字符串的每个字母代表一种数字。问多少种的等长的字符串通过&操作得到s。

思路：1&0=0,0&1=0,0&0=0，1&1=1。字符串里面的每一个字符ch的每一位与1 &操作，如果得到0就有3就情况，如果得到1只有一种情况。[0,63]的2进制最多有6位。

代码：

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
char s[100100];
int getidx(char c)
{
    if(c>='0'&&c<='9') return c-'0';
    if(c>='A'&&c<='Z') return c-'A'+10;
    if(c>='a'&&c<='z') return c-'a'+36;
    if(c=='-') return 62;
    if(c=='_') return 63;
}
int main()
{
    int i,j;
    scanf("%s",s);
    __int64 ans = 1;
    for(i=0; i<strlen(s); i++)
    {
        int p = getidx(s[i]);
        for(j=0; j<6; j++)
            if(((p>>j)&1)==0) ans=ans*3%mod;
    }
    cout<<ans<<endl;
}