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>
#pragma GCC optimize("O2")
using namespace std;
string s;
long long ss[80];
bool viz[80];
long long sol = 1;
int findnr(char x) {
  if (x == '-') return 62;
  if (x == '_') return 63;
  if (x <= '9') return (x - '0');
  if (x <= 'Z') return 10 + (x - 'A');
  if (x <= 'z') return 36 + (x - 'a');
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    int q = findnr(s[i]);
    if (viz[q])
      sol = (sol * ss[q]) % 1000000007;
    else {
      for (int a = 0; a <= 63; ++a)
        for (int b = 0; b <= 63; ++b) {
          int x = (a & b);
          if (x == q) ++ss[q];
        }
      viz[q] = 1;
      sol = (sol * ss[q]) % 1000000007;
    }
  }
  cout << sol;
  return 0;
}