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>
int N = 1000000007;
const int M = 100005;
int word[M];
int parse[255];
void initParse() {
  for (unsigned i = (65); i <= (90); i++) {
    parse[i] = i - 55;
  }
  for (unsigned i = (97); i <= (122); i++) {
    parse[i] = i - 61;
  }
  for (unsigned i = (48); i <= (57); i++) {
    parse[i] = i - 48;
  }
  parse[95] = 63;
  parse[45] = 62;
}
using namespace std;
int main() {
  initParse();
  long long res = 1;
  int n = 0;
  char c;
  while (cin >> c) {
    word[n++] = parse[(int)c];
  }
  for (unsigned i = (0); i <= (n - 1); i++) {
    int a = word[i];
    a <<= 1;
    for (unsigned i = (1); i <= (6); i++) {
      a >>= 1;
      int flag = a % 2;
      if (!flag) {
        res *= (long long)3;
        if (res > N) {
          res = res % N;
        }
      }
    }
  }
  cout << res;
  return 0;
}
