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.*;



public class Main {
		static public int Convert (char a){
					if ('0' <= a && a <= '9')
								return a - '0';
					if ('A' <= a && a <= 'Z')
								return a - 'A' + 10;
					if ('a' <= a && a <= 'z')
								return a - 'a' + 36;
					if (a == '-')
						return 62;
					if (a == '_')
						return 63;
					return 0;
					
		}
		public static void main (String args[]){
				Scanner in = new Scanner (System.in);
				String s = in.nextLine ();
				int n = s.length();
				int a[] = new int [n];
				int dp[] = new int [64];
														
				for (int i = 0 ; i < n ; ++i) {
							a[i] = Convert (s.charAt(i)); 
				} 
				
				for (int i = 0 ; i < 64 ; ++i) {
						for (int j = 0 ; j < 64 ; ++j) {
								dp[i & j] ++;
						}
				}
				final int mod = 1000000007;
				long ans = 1;
				
				for (int i = 0 ; i < n ; ++i) {
						ans = ans * dp[a[i]];
						ans %= mod;
												
				}
				System.out.println (ans);
				
													
		}
}