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 ProblemC {
	static long mod = 1_000_000_007;
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		String in = input.nextLine();
		long mult = 1;
		for(int a = 0; a < in.length(); a++){
			int bits = getNum0Bits(charTo64(in.charAt(a)));
			for(int b = 0; b < bits; b++){
				mult = mult * 3;
				mult = mult % mod;
			}
		}
		System.out.println(mult);
	}
	//I hear there is super fancy way to do this with bit stuff. Handbook? (or it's there already and I forgot)
	public static int getNum0Bits(int i){ 
		int bits = 0;
		for(int a =0; a < 6; a++){
			bits += 1-(i%2);
			i/=2;
		}
		return bits;
	}
	public static int charTo64(char i){
		if(i >= '0' && i <= '9'){
			return i-'0'+0;
		}
		if(i >= 'A' && i <= 'Z'){
			return i-'A'+10;
		}
		if(i >= 'a' && i <= 'z'){
			return i-'a'+36;
		}
		if(i == '-'){
			return 62;
		}
		return 63;
	}
}
