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;

/**
 * Created by user on 01.06.2016.
 */
public class C {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.next();
        long ans = 1;
        for (int i = 0; i < s.length(); i++) {
            int a = 0;
            int sc = 0;
            if (s.charAt(i) == '-') {
                a = 62;
            } else if (s.charAt(i) == '_') {
                a = 63;
            } else if (s.charAt(i) - '0' < 10) {
                a = s.charAt(i) - '0';
            } else if (s.charAt(i) - 'a' < 0) {
                a = s.charAt(i) - 'A' + 10;
            } else {
                a = s.charAt(i) - 'a' + 36;
            }
            for (int j = 0; j < 64; j++) {
                for (int k = 0; k < 64; k++) {
                    if ((j & k) == a) {
                        sc++;
                    }
                }
            }
            ans = (ans * sc) % 1000000007;
        }
        System.out.println(ans);
    }
}