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;
int f(char ch) {
  if (ch >= '0' && ch <= '9') return ch - '0';
  if (ch >= 'A' && ch <= 'Z') return ch - 'A' + 10;
  if (ch >= 'a' && ch <= 'z') return ch - 'a' + 36;
  if (ch == '_') return 63;
  if (ch == '-') return 62;
}
long long g(int num) {
  vector<int> vec(6, 0);
  for (int i = 0; i < 6; i++) {
    vec[i] = num % 2;
    num /= 2;
  }
  long long ret = 1;
  for (int i = 0; i < 6; i++) {
    if (vec[i] == 1)
      continue;
    else
      ret *= 3;
  }
  return ret;
}
void norm(long long &x) {
  if (x < 0) {
    x %= (long long)(1e9 + 7);
    x += (long long)(1e9 + 7);
  }
  while (x >= (long long)(1e9 + 7)) x %= (long long)(1e9 + 7);
}
int main() {
  string s;
  cin >> s;
  long long ans = 1;
  for (int i = 0; i < s.length(); i++) {
    ans = ans * (g(f(s[i])));
    norm(ans);
  }
  cout << ans;
}
