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 conv(char a) {
  if (a >= '0' && a <= '9') {
    return a - '0';
  } else if (a >= 'A' && a <= 'Z') {
    return a - 'A' + 10;
  } else if (a >= 'a' && a <= 'z') {
    return a - 'a' + 36;
  }
  if (a == '-') return 62;
  return 63;
}
int main() {
  string s;
  cin >> s;
  int tows = 0;
  int mx = 0;
  long long ans1 = 1;
  for (int i = 0; i < s.size(); i++) {
    mx = max(mx, conv(s[i]));
  }
  for (int i = 0; i < s.size(); i++) {
    tows = 0;
    long long ans = 1;
    int k = conv(s[i]);
    int co = 1;
    int it = 0;
    while (63 >= co) {
      if ((k >> it) % 2 == 0) tows++;
      it++;
      co *= 2;
    }
    for (int i = 0; i < tows; i++) {
      ans *= 3;
      ans %= 1000000007;
    }
    ans1 = ans1 % 1000000007 * ans % 1000000007;
    ans1 %= 1000000007;
  }
  ans1 %= 1000000007;
  cout << ans1 << endl;
}
