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;
const long long M = 1000000007;
int main() {
  string s;
  long long ans = 1;
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    int x, a[6], b = 32;
    if ('0' <= s[i] && s[i] <= '9') {
      x = s[i] - '0';
    } else if ('A' <= s[i] && s[i] <= 'Z') {
      x = s[i] - 'A' + 10;
    } else if ('a' <= s[i] && s[i] <= 'z') {
      x = s[i] - 'a' + 36;
    } else if (s[i] == '-') {
      x = 62;
    } else {
      x = 63;
    }
    for (int j = 0; j < 6; ++j) {
      a[j] = x / b;
      x %= b;
      b /= 2;
      if (a[j] == 0) {
        ans = (3 * ans) % M;
      }
    }
  }
  cout << ans << endl;
}