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 Get(char c) {
  if (c >= '0' && c <= '9') {
    return c - '0';
  }
  if (c >= 'A' && c <= 'Z') {
    return c - 'A' + 10;
  }
  if (c >= 'a' && c <= 'z') {
    return c - 'a' + 36;
  }
  if (c == '-') return 62;
  assert(c == '_');
  return 63;
}
long long M = 1000000000 + 7;
long long GetN(char c) {
  int n = Get(c);
  long long res = 0;
  for (int i = 0; i < 6; ++i) {
    if (((n >> i) & 1) == 0) {
      ++res;
    }
  }
  return res;
}
int main(int argc, char* argv[]) {
  std::ios::sync_with_stdio(false);
  string str;
  cin >> str;
  long long zero = 0;
  for (int i = 0; i < str.size(); ++i) {
    zero += GetN(str[i]);
  }
  long long res = 1;
  for (int i = 0; i < zero; ++i) res = (res * 3) % M;
  cout << res << endl;
  return 0;
}