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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class C {
	
	static long count_zeros(int x){
		String s = Integer.toBinaryString(x);
		long result = 6L;
		for (int i = 0; i < s.length(); i++) {
			if(s.charAt(i)=='1')
				result--;
		}
		return result;
	}
	static long power(long b, long p){
		long r = 1L;
		for (long i = 0L; i < p; i++) {
			r*=b;
		}
		return r;
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
			String in =br.readLine();
			long result = 1L;
			long pow=0L;
			for (int i = 0; i < in.length(); i++) {
				pow = 0L;
				if(in.charAt(i)>='0'&&in.charAt(i)<='9'){
					pow = count_zeros(in.charAt(i)-'0');
				}else if(in.charAt(i)>='A'&&in.charAt(i)<='Z'){
					pow = count_zeros(in.charAt(i)-'A'+10);
				}else if(in.charAt(i)>='a'&&in.charAt(i)<='z'){
					pow = count_zeros(in.charAt(i)-'a'+36);
				}else if (in.charAt(i)=='-'){
					pow = count_zeros(62);
				}else{
					pow = count_zeros(63);
				}
				
				pow= power(3L, pow);
				result*=pow;
				
				
			result=result %1000000007L;

				
			}
			System.out.println(result);
			
			
			
			
	}
}
