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;
string s;
long long int n, a, sum = 1, mod = 1000000007;
long long int to_ll(string x, long long int p) {
  long long int ans;
  long long int w = (long long int)x[p];
  if (47 < w && w < 58) {
    ans = w - 48;
  } else if (64 < w && w < 91) {
    ans = w - 65 + 10;
  } else if (96 < w && w < 123) {
    ans = w - 97 + 36;
  } else if (w == 95) {
    ans = 63;
  } else {
    ans = 62;
  }
  return ans;
}
long long int zero(long long int x) {
  long long int ans = 0;
  for (long long int i = 0; i < 6; i++) {
    if (x % 2 == 0) {
      ans++;
    }
    x /= 2;
  }
  return ans;
}
int main() {
  cin >> s;
  for (long long int i = 0; i < s.size(); i++) {
    long long int k = to_ll(s, i);
    long long int ze = zero(k);
    if (ze != 0) {
      sum *= (pow(3, ze));
    }
    sum = sum % mod;
  }
  cout << sum;
}