#include <bits/stdc++.h>
using namespace std;
multiset<int> data;
int getNumber(char c) {
  if (c >= '0' && c <= '9') {
    return c - '0';
  } else if (c >= 'A' && c <= 'Z') {
    return c - 'A' + 10;
  } else if (c >= 'a' && c <= 'z') {
    return c - 'a' + 36;
  } else if (c == '_') {
    return 63;
  } else {
    return 62;
  }
}
int main() {
  long long int n, h, k, i, j, ans, d;
  vector<char> lists;
  for (char c = '0'; c <= '9'; c++) {
    lists.push_back(c);
  }
  for (char c = 'A'; c <= 'Z'; c++) {
    lists.push_back(c);
  }
  for (char c = 'a'; c <= 'z'; c++) {
    lists.push_back(c);
  }
  lists.push_back('_');
  lists.push_back('-');
  string str;
  cin >> str;
  map<char, int> data;
  for (auto s : lists) {
    int curr = 0;
    for (auto c2 : lists) {
      for (auto c1 : lists) {
        if ((getNumber(c1) & getNumber(c2)) == getNumber(s)) {
          curr++;
        }
      }
    }
    data[s] = curr;
  }
  ans = 1;
  for (auto s : str) {
    ans *= data[s];
    if ((ans) >= 1000000007) ans %= 1000000007;
    ;
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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;
multiset<int> data;
int getNumber(char c) {
  if (c >= '0' && c <= '9') {
    return c - '0';
  } else if (c >= 'A' && c <= 'Z') {
    return c - 'A' + 10;
  } else if (c >= 'a' && c <= 'z') {
    return c - 'a' + 36;
  } else if (c == '_') {
    return 63;
  } else {
    return 62;
  }
}
int main() {
  long long int n, h, k, i, j, ans, d;
  vector<char> lists;
  for (char c = '0'; c <= '9'; c++) {
    lists.push_back(c);
  }
  for (char c = 'A'; c <= 'Z'; c++) {
    lists.push_back(c);
  }
  for (char c = 'a'; c <= 'z'; c++) {
    lists.push_back(c);
  }
  lists.push_back('_');
  lists.push_back('-');
  string str;
  cin >> str;
  map<char, int> data;
  for (auto s : lists) {
    int curr = 0;
    for (auto c2 : lists) {
      for (auto c1 : lists) {
        if ((getNumber(c1) & getNumber(c2)) == getNumber(s)) {
          curr++;
        }
      }
    }
    data[s] = curr;
  }
  ans = 1;
  for (auto s : str) {
    ans *= data[s];
    if ((ans) >= 1000000007) ans %= 1000000007;
    ;
  }
  cout << ans;
  return 0;
}

```