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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(0);
  vector<int> cnt(64, 0);
  for (int i = 0; i < 64; ++i)
    for (int j = 0; j < 64; ++j) ++cnt[i & j];
  string str;
  cin >> str;
  int l = str.length();
  long long ans = 1;
  for (int i = 0; i < l; ++i) {
    int t;
    if (str[i] == '-')
      t = 62;
    else if (str[i] == '_')
      t = 63;
    else if (str[i] >= 'A' && str[i] <= 'Z')
      t = str[i] - 'A' + 10;
    else if (str[i] >= 'a' && str[i] <= 'z')
      t = str[i] - 'a' + 36;
    else
      t = str[i] - '0';
    ans = (ans * cnt[t]) % (1000000007);
  }
  { std::cout << ans << '\n'; };
  return 0;
}