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);
  string s;
  string pattern;
  long long info[305];
  cin >> s;
  int n = s.length();
  long long ans = 1;
  for (char i = '0'; i <= '9'; i++) pattern.push_back(i);
  for (char i = 'A'; i <= 'Z'; i++) pattern.push_back(i);
  for (char i = 'a'; i <= 'z'; i++) pattern.push_back(i);
  pattern.push_back('-');
  pattern.push_back('_');
  for (int i = 0; i < 64; i++) info[pattern[i]] = i;
  for (int i = 0; i < n; i++) {
    int x = info[s[i]];
    for (int j = 0; j < 6; j++)
      if ((x & (1 << j)) == 0) ans = (ans * 3) % 1000000007;
  }
  cout << ans;
  return 0;
}