#include <bits/stdc++.h>
using namespace std;
inline void initialize_m(map<char, int> &m) {
  for (int i = 0; i < 10; ++i) m[i + '0'] = i;
  for (int i = 0; i < 26; ++i) {
    m['A' + i] = 10 + i;
    m['a' + i] = 36 + i;
  }
  m['-'] = 62;
  m['_'] = 63;
}
inline void init_zeros(int *zeros) {
  for (int i = 0; i < 64; ++i) zeros[i] = 0;
  for (int i = 0; i < 64; ++i) {
    if ((i / 32) % 2 == 0) zeros[i] += 1;
    if ((i / 16) % 2 == 0) zeros[i] += 1;
    if ((i / 8) % 2 == 0) zeros[i] += 1;
    if ((i / 4) % 2 == 0) zeros[i] += 1;
    if ((i / 2) % 2 == 0) zeros[i] += 1;
    if (i % 2 == 0) zeros[i] += 1;
  }
}
int main(int argc, const char *argv[]) {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  map<char, int> m;
  initialize_m(m);
  int zeros[64];
  init_zeros(zeros);
  int nways[7];
  int pows2[7];
  pows2[6] = 128;
  pows2[0] = 2;
  pows2[1] = 4;
  pows2[2] = 8;
  pows2[3] = 16;
  pows2[4] = 32;
  pows2[5] = 64;
  nways[0] = 1;
  nways[1] = 3;
  nways[2] = 9;
  nways[3] = 27;
  nways[4] = 81;
  nways[5] = 243;
  nways[6] = 729;
  vector<int> times;
  times.resize(s.length());
  for (int i = 0; i < s.length(); ++i) times[i] = nways[zeros[m[s[i]]]];
  long long res = times[0];
  for (int i = 1; i < s.length(); ++i)
    res = (res * (long long)times[i]) % 1000000007;
  cout << res << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
inline void initialize_m(map<char, int> &m) {
  for (int i = 0; i < 10; ++i) m[i + '0'] = i;
  for (int i = 0; i < 26; ++i) {
    m['A' + i] = 10 + i;
    m['a' + i] = 36 + i;
  }
  m['-'] = 62;
  m['_'] = 63;
}
inline void init_zeros(int *zeros) {
  for (int i = 0; i < 64; ++i) zeros[i] = 0;
  for (int i = 0; i < 64; ++i) {
    if ((i / 32) % 2 == 0) zeros[i] += 1;
    if ((i / 16) % 2 == 0) zeros[i] += 1;
    if ((i / 8) % 2 == 0) zeros[i] += 1;
    if ((i / 4) % 2 == 0) zeros[i] += 1;
    if ((i / 2) % 2 == 0) zeros[i] += 1;
    if (i % 2 == 0) zeros[i] += 1;
  }
}
int main(int argc, const char *argv[]) {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  map<char, int> m;
  initialize_m(m);
  int zeros[64];
  init_zeros(zeros);
  int nways[7];
  int pows2[7];
  pows2[6] = 128;
  pows2[0] = 2;
  pows2[1] = 4;
  pows2[2] = 8;
  pows2[3] = 16;
  pows2[4] = 32;
  pows2[5] = 64;
  nways[0] = 1;
  nways[1] = 3;
  nways[2] = 9;
  nways[3] = 27;
  nways[4] = 81;
  nways[5] = 243;
  nways[6] = 729;
  vector<int> times;
  times.resize(s.length());
  for (int i = 0; i < s.length(); ++i) times[i] = nways[zeros[m[s[i]]]];
  long long res = times[0];
  for (int i = 1; i < s.length(); ++i)
    res = (res * (long long)times[i]) % 1000000007;
  cout << res << endl;
  return 0;
}

```