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.Scanner;
import java.util.*;

/**
 *
 * @author IT manager
 */
public class JavaApplication1 {

    /**
     * @param args the command line arguments
     */
      public static int MOD=1000000007;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
         int [] freq= new int [64];
       String s = scanner.nextLine();// number of  potetos
       
    for(int i=0; i<64;i++){
    for(int j=0; j<64;j++){
            freq[i&j]++;
    }
    
    }
  
    long result=1;
    for (int i=0; i<s.length();i++){
        char x=s.charAt(i);
        int dig = 0;
        if(x >= '0' && x <= '9'){
        dig=s.charAt(i)-'0';
        }
         else if(x >= 'A' && x <= 'Z'){
        dig=s.charAt(i)-'A'+10 ;
        }
         else if(x >= 'a' && x<= 'z'){
        dig=s.charAt(i)-'a'+36;
        }
        else if(x == '-'){
        dig=62;
        }
        else {
        dig=63;
        }
        result=(result*freq[dig]) % MOD;
    }
    
      System.out.println(result);

  

}
}