#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int cnt[70];
long long int pow_mod(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  long long int temp = pow_mod(a, b / 2, m);
  temp = (temp * temp) % m;
  if (b & 1) temp = (temp * a) % m;
  return temp;
}
int main() {
  cnt[0] = 6;
  for (int i = 1; i <= 64; i++) {
    int num = i, cnts = 0;
    for (int j = 0; j < 6; j++) {
      if (!(num >> j & 1)) cnts++;
    }
    cnt[i] = cnts;
  }
  long long int tc = 0;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    char x = s[i];
    int y;
    if (isdigit(x)) {
      y = x - '0';
      cerr << "y"
           << ": " << y << endl;
      ;
    } else if (isupper(x)) {
      y = x - 'A' + 10;
      cerr << "y"
           << ": " << y << endl;
      ;
    } else if (x == '-') {
      y = 62;
    } else if (x == '_') {
      y = 63;
    } else {
      y = x - 'a' + 36;
    }
    tc += cnt[y];
  }
  tc = pow_mod(3, tc, MOD);
  cout << tc << endl;
}

### Prompt

Please create a solution in CPP 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;
const int MOD = 1000000007;
int cnt[70];
long long int pow_mod(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  long long int temp = pow_mod(a, b / 2, m);
  temp = (temp * temp) % m;
  if (b & 1) temp = (temp * a) % m;
  return temp;
}
int main() {
  cnt[0] = 6;
  for (int i = 1; i <= 64; i++) {
    int num = i, cnts = 0;
    for (int j = 0; j < 6; j++) {
      if (!(num >> j & 1)) cnts++;
    }
    cnt[i] = cnts;
  }
  long long int tc = 0;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    char x = s[i];
    int y;
    if (isdigit(x)) {
      y = x - '0';
      cerr << "y"
           << ": " << y << endl;
      ;
    } else if (isupper(x)) {
      y = x - 'A' + 10;
      cerr << "y"
           << ": " << y << endl;
      ;
    } else if (x == '-') {
      y = 62;
    } else if (x == '_') {
      y = 63;
    } else {
      y = x - 'a' + 36;
    }
    tc += cnt[y];
  }
  tc = pow_mod(3, tc, MOD);
  cout << tc << endl;
}

```