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>
using namespace std;
int n, m, sum, res, flag;
char s[1123456];
int ci[20];
void init() {
  ci[0] = 1;
  for (int i = 1; i < 10; i++) {
    ci[i] = ci[i - 1] * 3;
  }
}
int main() {
  init();
  int i, j, k, cas, T, t, x, y, z;
  while (scanf("%s", &s) != EOF) {
    n = strlen(s);
    long long res = 1;
    for (i = 0; i < n; i++) {
      if (s[i] == '-')
        t = 62;
      else if (s[i] == '_')
        t = 63;
      else if (s[i] >= '0' && s[i] <= '9')
        t = s[i] - '0';
      else if (s[i] >= 'a' && s[i] <= 'z')
        t = s[i] - 'a' + 36;
      else if (s[i] >= 'A' && s[i] <= 'Z')
        t = s[i] - 'A' + 10;
      z = 0;
      for (j = 0; j < 6; j++) {
        if (!(t & 1)) z++;
        t >>= 1;
      }
      res *= ci[z];
      res %= 1000000007;
    }
    printf("%I64d\n", res);
  }
  return 0;
}