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 calc(long long i) {
  int ans = 0;
  for (int j = 0; j < 6; j++) {
    if (!((1 << j) & i)) ans++;
  }
  return ans;
}
int decode(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 62;
  return 63;
}
string s;
long long sz, zeros[100], pos[100];
long long ans;
int main() {
  cin >> s;
  sz = s.size();
  ans = 1;
  for (int i = 0; i < 64; i++) {
    zeros[i] = calc(i);
    pos[i] = 1;
    for (int j = 0; j < zeros[i]; j++) {
      pos[i] *= 3;
      pos[i] %= (int)(1e9 + 7);
    }
  }
  for (int i = 0; i < sz; i++) {
    int n = decode(s[i]);
    ans *= pos[n];
    ans %= (int)(1e9 + 7);
  }
  cout << ans << endl;
}
