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 z[70];
int main() {
  string s;
  cin >> s;
  for (int i = 0; i <= 63; i++) {
    int temp = i;
    int c = 0, in = 0;
    while (temp) {
      if (temp % 2 == 0) c++;
      temp /= 2;
      in++;
    }
    z[i] = c + (6 - in);
  }
  map<char, int> m;
  m['0'] = 0;
  m['1'] = 1;
  m['2'] = 2;
  m['3'] = 3;
  m['4'] = 4;
  m['5'] = 5;
  m['6'] = 6;
  m['7'] = 7;
  m['8'] = 8;
  m['9'] = 9;
  int c = 10;
  for (char i = 'A'; i <= 'Z'; i++) {
    m[i] = c;
    c++;
  }
  c = 36;
  for (char i = 'a'; i <= 'z'; i++) {
    m[i] = c;
    c++;
  }
  m['-'] = 62;
  m['_'] = 63;
  long long ans = 1;
  int M = 1000000007;
  for (int i = 0; i < s.length(); i++) {
    int temp = z[m[s[i]]];
    while (temp) {
      ans = (ans * 3) % M;
      temp--;
    }
  }
  cout << ans << endl;
  return 0;
}
