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;
long long m = 1000000007;
long long bigmod(long long b, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long t;
    t = bigmod(3, p / 2);
    return ((t % m) * (t % m)) % m;
  } else {
    long long t = bigmod(3, p - 1);
    return (b * (t % m)) % m;
  }
}
int main() {
  string s;
  while (cin >> s) {
    long long num = 0, c = 0;
    for (long long i = 0; i < s.length(); i++) {
      if (s[i] == '-')
        num = 62;
      else if (s[i] == '_')
        num = 63;
      else if ((int)(s[i]) >= 48 && (int)s[i] <= 57)
        num = (int)s[i] - 48;
      else if ((int)s[i] >= 65 && (int)s[i] <= 90)
        num = (int)s[i] - 65 + 10;
      else
        num = (int)s[i] - 97 + 36;
      for (long long j = 0; j < 6; j++) {
        long long t = ceil(pow(2, j));
        long long ret = (t & num);
        if (ret == 0) c++;
      }
    }
    cout << bigmod(3, c) << endl;
  }
  return 0;
}