#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
int getint(char s) {
  if (s >= '0' && s <= '9') {
    return (s - '0');
  }
  if (s >= 'A' && s <= 'Z') {
    return (s - 'A') + 10;
  }
  if (s >= 'a' && s <= 'z') {
    return (s - 'a') + 36;
  }
  if (s == '-') {
    return 62;
  }
  return 63;
}
int get0(int num) {
  int i = 6;
  int count = 0;
  while (i != 0) {
    if (!(num & 1)) {
      count++;
    }
    num >>= 1;
    i--;
  }
  return count;
}
int getans(string s) {
  long long count = 1;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    int num = getint(s[i]);
    int num0 = get0(num);
    count = (count * pow(3, num0));
    count = count % mod;
  }
  return count;
}
int main() {
  string s;
  cin >> s;
  cout << getans(s);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
int getint(char s) {
  if (s >= '0' && s <= '9') {
    return (s - '0');
  }
  if (s >= 'A' && s <= 'Z') {
    return (s - 'A') + 10;
  }
  if (s >= 'a' && s <= 'z') {
    return (s - 'a') + 36;
  }
  if (s == '-') {
    return 62;
  }
  return 63;
}
int get0(int num) {
  int i = 6;
  int count = 0;
  while (i != 0) {
    if (!(num & 1)) {
      count++;
    }
    num >>= 1;
    i--;
  }
  return count;
}
int getans(string s) {
  long long count = 1;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    int num = getint(s[i]);
    int num0 = get0(num);
    count = (count * pow(3, num0));
    count = count % mod;
  }
  return count;
}
int main() {
  string s;
  cin >> s;
  cout << getans(s);
  return 0;
}

```