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.HashMap;
import java.util.Scanner;

public class R355Div2C {
	static int MOD=1_000_000_007;
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		String s=in.next();
		char[] c="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_".toCharArray();
		HashMap<Character,Integer> map=new HashMap<Character,Integer>();
		for(int i=0;i<64;i++)
			map.put(c[i],0);
		
		for(int i=0;i<64;i++){
			for(int j=0;j<64;j++){
				if(i!=j){
					int x=i&j;
					map.put(c[x],map.get(c[x])+1);
				}
			}
		}
		
		int n=s.length();
		long ans=1;
		for(int i=0;i<n;i++){
			long x=map.get(s.charAt(i));
			ans=ans*(x+1)%MOD;
		}
		System.out.println(ans);
	}
}
