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 scanner = new Scanner( System.in );
		int[] lookup = new int[64];
		for( int i = 1; i < 64; ++i ){
			if( ( i & 1 ) == 1 ) lookup[i] = lookup[i - 1] + 1;
			else lookup[i] = lookup[i >> 1];
		}
		for( int i = 0; i < 64; ++i ) lookup[i] = 6 - lookup[i];
		int cnt = scanner.next( ).chars( ).map( c -> {
			if( c == '_' ) c = 63;
			else if( c == '-' ) c = 62;
			else if( c <= '9' ) c = c - '0';
			else if( c < 'a' ) c = c - 'A' + 10;
			else c = c - 'a' + 36;
			return lookup[c];
		} ).sum( );
		long res = 1, base = 3;
		while( cnt > 0 ){
			if( ( cnt & 1 ) == 0 ){
				base = base * base % 1_000_000_007;
				cnt >>= 1;
			}else{
				res = res * base % 1_000_000_007;
				--cnt;
			}
		}
		System.out.println( res );
	}
}