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
import java.util.Scanner;
import java.util.Arrays;
import java.util.HashMap;
import java.lang.Math;

public class Codeforces{

	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		String str = in.next();
		long max = (long)Math.pow(10,9) + 7;
		long count = 1;
		StringBuilder sb = new StringBuilder();
		for(char b : str.toCharArray()){
			int c = (int)b;
			if(c >= 48 && c <= 57){
				c -= 48; 
			} else if(c >= 65 && c <= 90){
				c -= 55;
			} else if(c >= 97 && c <= 122){
				c -= 61;
			} else if(b == '-'){
				c = 62;
			} else if(b == '_'){
				c = 63;
			}
			String s = String.format("%6s", Integer.toBinaryString(c)).replace(' ', '0');
			for(char k : s.toCharArray()){
				if(k == '1'){
					
				} else {
					count *= 3;
				}
				count = count % max;
			}
		}
		System.out.println(count);
	}
}