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;


public class C {
public static void main(String[] args){
	Scanner in = new Scanner (System.in);
	
	String word = in.next();
	
	long count = 1;
	
	for (char c : word.toCharArray()){
		for (int i = 0; i < Integer.bitCount((~convert(c))&63); i++)
		count*=3;
		count%=1000000007;
	}
	System.out.println(count);
	
	
	
}
static int convert(char c){
	if (c >= '0' && c <= '9') return c-'0';
	if (c >= 'A' && c <= 'Z') return c-'A'+10;
	if (c >= 'a' && c <= 'z') return c-'a'+36;
	if (c == '-') return 62;
	return 63;
}
}
