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>
using namespace std;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long temp = power(x, int(y / 2));
  if (y % 2 == 0)
    return (((temp % 1000000007) * temp) % 1000000007);
  else
    return ((((x * temp) % 1000000007) * temp) % 1000000007);
}
int countsetbit(int n) {
  unsigned int count = 0;
  while (n) {
    n &= (n - 1);
    count++;
  }
  return count;
}
int main() {
  int mat[257];
  for (long long i = ((int)'0'); i <= (int)'9'; i++) mat[i] = i - (int)'0';
  for (long long i = ((int)'a'); i <= (int)'z'; i++) {
    mat[i] = 36 + i - (int)'a';
  }
  for (long long i = ((int)'A'); i <= (int)'Z'; i++) {
    mat[i] = 10 + i - (int)'A';
  }
  mat[(int)'-'] = 62;
  mat[(int)'_'] = 63;
  string s;
  cin >> s;
  long long c = 0;
  for (long long i = 0; i <= s.length() - 1; i++) {
    c += 6 - countsetbit(mat[s[i]]);
  }
  cout << power(3, c) << endl;
}
