Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Class{
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine();
        long product = 1;
        for(int i=0; i<s.length(); i++){
            char c = s.charAt(i);
            long x = 0;
            if(c>='0'&&c<='9')x=c-'0';
            else if(c>='A'&&c<='Z')x=c-'A'+10;
            else if(c>='a'&&c<='z')x=c-'a'+36;
            else if(c=='-')x=62;
            else if(c=='_')x=63;
            int y = count(x);
            product = (product*((long)Math.pow(3,y)))%(1000000007);
        }
        System.out.print((product)%(1000000007));
    }
    
    public static int count(long x){
        int count = 0;
        for(int i=0; i<6; i++){
            count+=x%2;
            x=x/2;
        }
        return 6-count;
    }
}