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.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class BB
{
   
public static void main(String[] args)
{
    
    Scanner sc= new Scanner(System.in);
    
    
    
    String s=sc.next();
    int l=s.length();
    int sl;
    long num=1;
   
    for (int i = 0; i < l; i++) {
         long cnum=0;
        if(s.charAt(i)>='0'&&s.charAt(i)<='9')
            sl=s.charAt(i)-'0';
        else   if(s.charAt(i)>='A'&&s.charAt(i)<='Z'){
            sl=s.charAt(i)-'A'+10;
        }
        else   if(s.charAt(i)>='a'&&s.charAt(i)<='z')
        {
             sl=s.charAt(i)-'a'+36;
        }
        else
            if(s.charAt(i)=='-')
                sl=62;
        else
                sl=63;
        
     
        for (int j = 0; j < 64; j++) {
            for (int k = 0; k < 64; k++) {
                 if((j&k)==sl)
                cnum++;
            }
           
        }
        num*=cnum;
        num=num%1000000007;
    }
   
   
    System.out.println(num);
}



}