Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  long long i, l, ans, num, j;
  long long arr[100];
  char c;
  char str[100010];
  scanf("%s", str);
  l = strlen(str);
  ans = 1;
  for (i = 0; i < 64; i++) arr[i] = 0;
  for (i = 0; i < 64; i++) {
    for (j = 0; j < 64; j++) {
      num = (i & j);
      arr[num]++;
    }
  }
  for (i = 0; i < l; i++) {
    c = str[i];
    if (c == '-')
      num = 62;
    else if (c == '_')
      num = 63;
    else if (c >= 'a' && c <= 'z')
      num = (c - 'a') + 36;
    else if (c >= '0' && c <= '9')
      num = (c - '0');
    else
      num = c - 'A' + 10;
    ans = ans * arr[num];
    ans %= 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}