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


public class Tmp {
	public static void main(String[] args) throws Throwable
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));		
		String s = br.readLine();
		HashMap<Character, Integer> map = new HashMap<Character, Integer>();
		for (int i = 48; i <58; i++) 
		{
			map.put((char)i, i-48);
		}
		for (int i = 65; i < 91; i++) 
		{
			map.put((char)i,(i-'A')+10 );
		}
		for (int i = 97; i <123; i++) 
		{
			map.put((char)i, i-61);
		}
		map.put('-', 62);
		map.put('_',63);
		long ans = 1;
		long mod = (long) (1e9)+7;
		for (int i = 0; i < s.length(); i++) 
		{
//			String cur = Long.toBinaryString(map.get(s.charAt(i)));
//			ans += (long) (3*(64-cur.length())%mod);
			long mask = 1;
			int x = map.get(s.charAt(i));
			for(int j = 0;j<6;j++,mask<<=1)
			{
//				if(cur.charAt(j)=='0')
//					ans *= (long) 3%mod;
//				else
//					ans *= (long) 1%mod;
				if((x&mask)==0)
					{
						ans %= mod;
						ans *= (long) (3)%mod;
					}
			}
		}
		System.out.println(ans%mod);
	}
}
