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;

public class VanyaAndLabel {
    
    private static final long MOD = (long) Math.pow(10, 9) + 7;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        String s = scanner.nextLine();

        System.out.println(solve(s));
    }
    
    private static long solve(String s) {
        long count = 1;

        for (int i = 0; i < s.length(); i++) {

            char c = s.charAt(i);
        
            int x;

            if (between(c, '0', '9'))
                x = c - '0';
            else if (between(c, 'A', 'Z'))
                x = c - 'A' + 10;
            else if (between(c, 'a', 'z'))
                x = c - 'a' + 36;
            else if (c == '-')
                x = 62;
            else
                x = 63;

            for (int j = 0; j < 6; j++)
                if (((x >> j) & 1) == 0)
                    count = Math.floorMod(count * 3, MOD);

        }

        return count;
    }
    
    private static boolean between(char c, char l, char r) {
        return c >= l && c <= r;
    }
}