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;
struct my_test {
  void operator()() {
    string str;
    cin >> str;
    int n = str.size();
    vector<int64_t> v(n);
    for (int i = 0; i < n; ++i) {
      if (str[i] >= '0' && str[i] <= '9') {
        v[i] = str[i] - '0';
      } else if (str[i] >= 'A' && str[i] <= 'Z') {
        v[i] = str[i] - 'A' + 10;
      } else if (str[i] >= 'a' && str[i] <= 'z') {
        v[i] = str[i] - 'a' + 36;
      } else if (str[i] == '-')
        v[i] = 62;
      else if (str[i] == '_')
        v[i] = 63;
    }
    int64_t ans = 1;
    int64_t mod = 1000 * 1000 * 1000 + 7;
    for (int i = 0; i < n; ++i) {
      int64_t cur = 0;
      for (int x = 0; x < 64; ++x) {
        for (int y = 0; y < 64; ++y) {
          if ((x & y) == v[i]) {
            ++cur;
          }
        }
      }
      ans = (ans * cur) % mod;
    }
    cout << ans % mod << endl;
  }
};
void solve(int num_test = 0) { my_test()(); }
int main(void) {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
