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.*;

public class Main {
  public static void main(String args[]) throws Exception {
    Scanner sc = new Scanner(System.in);
    String s = sc.next();
    int mod = 1000000007;

    int cnt = 0;

    for(char c : s.toCharArray()) {
      int bitNum = 0;

      if(Character.isDigit(c)) {
        bitNum = c - '0';
      } else if(Character.isLowerCase(c)) {
        bitNum = 36 + (c - 'a');
      } else if(Character.isUpperCase(c))
        bitNum = 10 + (c - 'A');
      else if(c == '-')
        bitNum = 62;
      else
        bitNum = 63;
      for(int i = 0; i < 6; i++) {
        if(((bitNum >> i) & 1) == 0) cnt++;
      }

    }
    long res = 1;
    while(cnt > 0) {
      res *= 3;
      res %= mod;
      cnt--;
    }
    System.out.println(res);
  }
}
