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;
const long long mod = (long long)1e9 + 7;
string second;
string p = "";
string binary(int a) {
  string r = "";
  for (int i = 5; i >= 0; i--)
    if (a & (1 << i))
      r += "1";
    else
      r += "0";
  return r;
}
string toBase(char c) {
  int r;
  if (c >= '0' && c <= '9') r = c - '0';
  if (c >= 'A' && c <= 'Z') r = c - 'A' + 10;
  if (c >= 'a' && c <= 'z') r = c - 'a' + 36;
  if (c == '-') r = 62;
  if (c == '_') r = 63;
  return binary(r);
}
int main() {
  cin >> second;
  for (int i = 0; i < (int)second.size(); i++) p += toBase(second[i]);
  long long ans = 1;
  for (int i = 0; i < (int)p.size(); i++)
    if (p[i] == '0') ans = (ans * 3) % mod;
  cout << ans << endl;
  return 0;
}