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;
string FileName = "minsumdiff";
string a;
int f(char a) {
  if ('0' <= a && a <= '9') return 0 + a - '0';
  if ('A' <= a && a <= 'Z') return 10 + a - 'A';
  if ('a' <= a && a <= 'z') return 36 + a - 'a';
  if (a == '-') return 62;
  return 63;
}
int cnt;
int Pow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b != 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> a;
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < 6; ++j) {
      if (f(a[i]) & (1 << j)) {
        --cnt;
      }
    }
    cnt += 6;
  }
  cout << Pow(3, cnt, 1e9 + 7);
}
