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 ::std::cin;
using ::std::cout;
using ::std::endl;
using ::std::string;
using ::std::unordered_map;
int main() {
  ::std::ios_base::sync_with_stdio(false);
  unordered_map<char, int> numbers;
  for (char c = '0'; c <= '9'; ++c) numbers[c] = c - '0';
  for (char c = 'A'; c <= 'Z'; ++c) numbers[c] = c - 'A' + 10;
  for (char c = 'a'; c <= 'z'; ++c) numbers[c] = c - 'a' + 36;
  numbers['-'] = 62;
  numbers['_'] = 63;
  const int MOD = 1e9 + 7;
  unordered_map<char, int> ways;
  for (auto charAndNumber : numbers) {
    ways[charAndNumber.first] = 1;
    for (int i = 0; i < 6 - __builtin_popcount(charAndNumber.second); ++i)
      ways[charAndNumber.first] = 3LL * ways[charAndNumber.first] % MOD;
  }
  string s;
  cin >> s;
  int result = 1;
  for (char c : s) result = 1LL * result * ways[c] % MOD;
  cout << result << endl;
  return 0;
}
