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 n, m, k;
long long pow(long long a, long long b) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > 1000000007) x %= 1000000007;
    }
    y = (y * y);
    if (y > 1000000007) y %= 1000000007;
    b /= 2;
  }
  return x;
}
int main() {
  string str;
  cin >> str;
  int ctr = 0;
  for (int i = 0; i < str.size(); i++) {
    int n;
    if (str[i] >= '0' && str[i] <= '9')
      n = str[i] - '0';
    else if (str[i] >= 'A' && str[i] <= 'Z')
      n = str[i] - 'A' + 10;
    else if (str[i] >= 'a' && str[i] <= 'z')
      n = str[i] - 'a' + 36;
    else if (str[i] == '-')
      n = 62;
    else
      n = 63;
    for (int i = 0; i < 6; i++) {
      if ((n >> i & 1) == 0) ctr++;
    }
  }
  cout << pow(3, (long long)ctr) << endl;
  return 0;
}
