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;
import java.util.StringTokenizer;

public class VanyaLabel {
    static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in), 32768);
    static StringTokenizer st;

    public static void main(String[] args) throws IOException {
        String s = bf.readLine();

        final double MOD = 1E9 + 7;

        int[] m = new int[1000000];
        for(int i = 0; i <= 9; i++) m['0' + i] = i;
        for(int i = 0; i < 26; i++) m['A' + i] = 10 + i;
        for(int i = 0; i < 26; i++) m['a' + i] = 36 + i;
        m['-'] = 62; m['_'] = 63;
        long ans = 1;

        for(int i = 0; i < s.length(); i++) {
            for (int j = 0; j < 6; j++)
                if ((m[s.charAt(i)] & (1 << j)) == 0)
                    ans = (long) ((ans * 3) % MOD);
        }

        System.out.println(ans);
    }
}
