#include <bits/stdc++.h>
using namespace std;
long long int modularexp(long long int base, long long int exp) {
  long long int res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000007;
    exp = exp >> 1;
    base = (base * base) % 1000000007;
  }
  return res;
}
int Get_Zeros(int n) {
  int i, c;
  c = 0;
  for (i = 1; i <= 6; i++) {
    if (n % 2 == 0) c++;
    n = n / 2;
  }
  return c;
}
int main() {
  int i, j, n, x;
  char str[100001];
  long long int z, ans;
  cin.getline(str, 100001, '\n');
  n = strlen(str);
  z = 0;
  for (i = 0; i < n; i++) {
    x = int(str[i]);
    if (x >= 65 && x <= 90)
      x -= 55;
    else if (x >= 97 && x <= 122)
      x -= 61;
    else if (x >= 48 && x <= 57)
      x -= 48;
    else if (x == 45)
      x = 62;
    else
      x = 63;
    z += Get_Zeros(x);
  }
  ans = modularexp(3, z);
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int modularexp(long long int base, long long int exp) {
  long long int res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000007;
    exp = exp >> 1;
    base = (base * base) % 1000000007;
  }
  return res;
}
int Get_Zeros(int n) {
  int i, c;
  c = 0;
  for (i = 1; i <= 6; i++) {
    if (n % 2 == 0) c++;
    n = n / 2;
  }
  return c;
}
int main() {
  int i, j, n, x;
  char str[100001];
  long long int z, ans;
  cin.getline(str, 100001, '\n');
  n = strlen(str);
  z = 0;
  for (i = 0; i < n; i++) {
    x = int(str[i]);
    if (x >= 65 && x <= 90)
      x -= 55;
    else if (x >= 97 && x <= 122)
      x -= 61;
    else if (x >= 48 && x <= 57)
      x -= 48;
    else if (x == 45)
      x = 62;
    else
      x = 63;
    z += Get_Zeros(x);
  }
  ans = modularexp(3, z);
  cout << ans << endl;
  return 0;
}

```