#include <bits/stdc++.h>
using namespace std;
const int p = 1000000007;
int pow3[10];
int main() {
  pow3[0] = 1;
  pow3[1] = 3;
  pow3[2] = pow3[1] * 3;
  pow3[3] = pow3[2] * 3;
  pow3[4] = pow3[3] * 3;
  pow3[5] = pow3[4] * 3;
  pow3[6] = pow3[5] * 3;
  map<char, int> myMap;
  char c;
  long long n, k = 0, cnt = 0, res = 1;
  for (int i = 48; i <= 57; i++) {
    c = i;
    myMap.insert(make_pair(c, k));
    k++;
  }
  for (int i = 65; i <= 90; i++) {
    c = i;
    myMap.insert(make_pair(c, k));
    k++;
  }
  for (int i = 97; i <= 122; i++) {
    c = i;
    myMap.insert(make_pair(c, k));
    k++;
  }
  myMap.insert(make_pair('-', 62));
  k++;
  myMap.insert(make_pair('_', 63));
  k++;
  while (cin >> c) {
    n = myMap[c];
    cnt = 0;
    for (int i = 0; i < 6; i++) {
      if (n % 2 == 0) cnt++;
      n /= 2;
    }
    res = ((res % p) * pow3[cnt]) % p;
  }
  cout << res;
  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;
const int p = 1000000007;
int pow3[10];
int main() {
  pow3[0] = 1;
  pow3[1] = 3;
  pow3[2] = pow3[1] * 3;
  pow3[3] = pow3[2] * 3;
  pow3[4] = pow3[3] * 3;
  pow3[5] = pow3[4] * 3;
  pow3[6] = pow3[5] * 3;
  map<char, int> myMap;
  char c;
  long long n, k = 0, cnt = 0, res = 1;
  for (int i = 48; i <= 57; i++) {
    c = i;
    myMap.insert(make_pair(c, k));
    k++;
  }
  for (int i = 65; i <= 90; i++) {
    c = i;
    myMap.insert(make_pair(c, k));
    k++;
  }
  for (int i = 97; i <= 122; i++) {
    c = i;
    myMap.insert(make_pair(c, k));
    k++;
  }
  myMap.insert(make_pair('-', 62));
  k++;
  myMap.insert(make_pair('_', 63));
  k++;
  while (cin >> c) {
    n = myMap[c];
    cnt = 0;
    for (int i = 0; i < 6; i++) {
      if (n % 2 == 0) cnt++;
      n /= 2;
    }
    res = ((res % p) * pow3[cnt]) % p;
  }
  cout << res;
  return 0;
}

```