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 c355;

import java.util.Arrays;
import java.util.Scanner;

public class q3 
{

	public static void main(String args[])
	{
		Scanner s=new Scanner(System.in);
		long mod=(long)(Math.pow(10, 9)+7);
		int[] pre=new int[64];
		for(int i=0;i<64;i++)
		{
			for(int j=0;j<64;j++)
			{
				pre[i&j]++;
			}
		}
	//	System.out.println(Arrays.toString(pre));
		String st=s.next();
		long ans=1;
		for(int i=0;i<st.length();i++)
		{
			int n;
			if((int)st.charAt(i)>='0' && (int)st.charAt(i)<='9' )
			{
				n=st.charAt(i)-'0';
				
			}else
			{
				if((int)st.charAt(i)>='A' && (int)st.charAt(i)<='Z' )
				{
					n=10+(st.charAt(i)-'A');
					
				}
				else
				{
					if((int)st.charAt(i)>='a' && (int)st.charAt(i)<='z' )
					{
						n=36+(st.charAt(i)-'a');
					}
					else
					{
						if(st.charAt(i)=='-')
							n=62;
						else
							n=63;
					}
				}
			}
		//	System.out.println(n);
			ans=(ans*pre[n])%mod;
		}
	    System.out.println(ans);	
	}
}