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;
const int MOD = 1000000007;
int f(string S) {
  map<char, int> bit;
  int size = S.size();
  int num = 0;
  int j;
  long long int count = 1;
  for (int i = 0; i < 64; i++) {
    if ((i >= 0) && (i < 10)) {
      bit[(char)(i + 48)] = i;
    }
    if ((i >= 10) && (i < 36)) {
      bit[(char)(i + 55)] = i;
    }
    if ((i >= 36) && (i < 62)) {
      bit[(char)(i + 61)] = i;
    }
    if (i == 62) {
      bit['-'] = i;
    } else {
      bit['_'] = i;
    }
  }
  for (int i = 0; i < size; i++) {
    j = bit[S.at(i)];
    for (int k = 0; k < 6; k++, j = j >> 1) {
      if (!(j & 1)) {
        num++;
      }
    }
  }
  for (int i = 0; i < num; i++) {
    count = (count * 3) % MOD;
  }
  return count;
}
int main() {
  string str;
  cin >> str;
  cout << f(str);
}
