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.Arrays;
import java.util.Scanner;

public class C {
	
	static String alpha = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_";
	static int mod = 1000000007;
	
	public static void main(String[] args) {
		Scanner reader = new Scanner(System.in);
		int[] sol = new int[alpha.length()];
		for(int i = 0; i < alpha.length(); i++) {
			for(int j = 0; j < alpha.length(); j++) {
				for(int k = 0; k < alpha.length(); k++) {
					if((j & k) == i) {
						sol[i]++;
					}
				}
			}	
		}
		
		String s = reader.nextLine();
		int[] a = new int[s.length()];
		for(int i = 0; i < s.length(); i++) {
			a[i] = alpha.indexOf(s.charAt(i));
		}
		long mul = 1;
		for(int i = 0; i < a.length; i++) {
			mul *= sol[a[i]];
			mul %= mod;
		}
		System.out.println(mul % mod);
		reader.close();
	}
	
}