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;
map<char, int> m;
int nopair[10000];
int main() {
  int i, j, k;
  for (i = 0; i <= 9; i++) {
    m['0' + i] = i;
  }
  j = 0;
  for (i = 'A'; i <= 'Z'; i++) {
    m[char(i)] = 10 + j;
    j++;
  }
  j = 0;
  for (i = 'a'; i <= 'z'; i++) {
    m[char(i)] = 36 + j;
    j++;
  }
  m['-'] = 62;
  m['_'] = 63;
  map<char, int>::iterator iter;
  for (iter = m.begin(); iter != m.end(); iter++) {
    char ch = iter->first;
    int val = iter->second;
    map<char, int>::iterator iter2;
    map<char, int>::iterator iter3;
    for (iter2 = m.begin(); iter2 != m.end(); iter2++) {
      for (iter3 = m.begin(); iter3 != m.end(); iter3++) {
        int andval = iter2->second & iter3->second;
        if (andval == iter->second) {
          nopair[val]++;
        }
      }
    }
  }
  string s;
  cin >> s;
  int n = s.size();
  long long totval = 1;
  for (i = 0; i < n; i++) {
    totval = (totval % 1000000007 * nopair[m[s[i]]] % 1000000007) % 1000000007;
  }
  cout << totval;
  return 0;
}