#include <bits/stdc++.h>
using namespace std;
int main() {
  long long myMod = 1000000007;
  string str = "";
  long long arr[64] = {0};
  unordered_map<char, int> mymap;
  vector<char> vec(64);
  for (int i = 0; i <= 9; i++) {
    char c = '0' + i;
    mymap[c] = i;
    vec[i] = c;
  }
  int cnt = 10;
  for (int i = 0; i < 26; i++) {
    char c = 'A' + i;
    mymap[c] = cnt;
    vec[cnt] = c;
    cnt++;
  }
  for (int i = 0; i < 26; i++) {
    char c = 'a' + i;
    mymap[c] = cnt;
    vec[cnt] = c;
    cnt++;
  }
  mymap['-'] = 62;
  mymap['_'] = 63;
  vec[62] = '-';
  vec[63] = '_';
  for (int i = 0; i < 64; i++) {
    for (int j = 0; j < 64; j++) {
      int a = mymap[vec[i]], b = mymap[vec[j]];
      int res = a & b;
      arr[res]++;
    }
  }
  cin >> str;
  long long result = 1;
  int sz = str.length();
  for (int k = 0; k < sz; k++) {
    long long a = (result % myMod), b = arr[mymap[str[k]]] % myMod;
    result = (a * b) % myMod;
  }
  cout << result << 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;
int main() {
  long long myMod = 1000000007;
  string str = "";
  long long arr[64] = {0};
  unordered_map<char, int> mymap;
  vector<char> vec(64);
  for (int i = 0; i <= 9; i++) {
    char c = '0' + i;
    mymap[c] = i;
    vec[i] = c;
  }
  int cnt = 10;
  for (int i = 0; i < 26; i++) {
    char c = 'A' + i;
    mymap[c] = cnt;
    vec[cnt] = c;
    cnt++;
  }
  for (int i = 0; i < 26; i++) {
    char c = 'a' + i;
    mymap[c] = cnt;
    vec[cnt] = c;
    cnt++;
  }
  mymap['-'] = 62;
  mymap['_'] = 63;
  vec[62] = '-';
  vec[63] = '_';
  for (int i = 0; i < 64; i++) {
    for (int j = 0; j < 64; j++) {
      int a = mymap[vec[i]], b = mymap[vec[j]];
      int res = a & b;
      arr[res]++;
    }
  }
  cin >> str;
  long long result = 1;
  int sz = str.length();
  for (int k = 0; k < sz; k++) {
    long long a = (result % myMod), b = arr[mymap[str[k]]] % myMod;
    result = (a * b) % myMod;
  }
  cout << result << endl;
  return 0;
}

```