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
#include <bits/stdc++.h>
int main() {
  int sum = 0;
  int tmp;
  unsigned long tmpresult;
  while ((tmp = getchar()) != '\n') {
    if (tmp == 45)
      tmp = 62;
    else if (tmp < 58)
      tmp -= 48;
    else if (tmp <= 90)
      tmp -= 55;
    else if (tmp == 95)
      tmp = 63;
    else
      tmp -= 61;
    for (int i = 0; i < 6; i++) {
      sum += ((tmp << (31 - i)) >> 31);
    }
    sum += 6;
  }
  tmpresult = 1;
  for (int i = 0; i < sum; i++) tmpresult = (tmpresult * 3) % 1000000007;
  printf("%lu", tmpresult);
  return 0;
}
