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:
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		long ans = 1;
		for(int i = 0;i<s.length();i++){
			char c = s.charAt(i);
			int r = (int)c;
			int k = 0;
			if((int)c>=(int)'0'&&(int)c<=(int)'0'+9){
				k = (int)c-(int)'0';
			}
			if(r - (int)'A'>=0&&r - (int)'A'<=25){
				k=r - (int)'A'+10;
			}
			if(r-(int)'a'>=0&&r-(int)'a'<=25){
				k= r-(int)'a'+36;
			}
			if(c=='-'){
				k=62;
			}
			if(c=='_'){
				k=63;
			}
			for(int j= 0;j<6;j++){
				if(k%2==0){
					ans=ans*3%(1000000007);
				}
				k/=2;
			}
		}
		System.out.println(ans);
	}
}