#include <bits/stdc++.h>
using namespace std;
template <typename T>
T absll(T X) {
  if (X < 0)
    return -1 * X;
  else
    return X;
}
long long modPow(long long A, long long B, long long C) {
  long long X = 1LL;
  while (B) {
    if (B & 1) {
      X = (X * A) % C;
    }
    A = (A * A) % C;
    B /= 2LL;
  }
  return X % C;
}
int main() {
  map<char, int> base;
  int val = 0;
  for (char ch = '0'; ch <= '9'; ch++) {
    base[ch] = val;
    ++val;
  }
  for (char ch = 'A'; ch <= 'Z'; ch++) {
    base[ch] = val;
    ++val;
  }
  for (char ch = 'a'; ch <= 'z'; ch++) {
    base[ch] = val;
    ++val;
  }
  base['-'] = 62;
  base['_'] = 63;
  string str;
  cin >> str;
  long long cnt = 0LL;
  for (int i = 0; i < str.length(); i++) {
    int num = base[str[i]];
    bitset<6> B(num);
    B.flip();
    cnt += (long long)B.count();
  }
  printf("%lld\n", modPow(3LL, cnt, 1000000007LL));
  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;
template <typename T>
T absll(T X) {
  if (X < 0)
    return -1 * X;
  else
    return X;
}
long long modPow(long long A, long long B, long long C) {
  long long X = 1LL;
  while (B) {
    if (B & 1) {
      X = (X * A) % C;
    }
    A = (A * A) % C;
    B /= 2LL;
  }
  return X % C;
}
int main() {
  map<char, int> base;
  int val = 0;
  for (char ch = '0'; ch <= '9'; ch++) {
    base[ch] = val;
    ++val;
  }
  for (char ch = 'A'; ch <= 'Z'; ch++) {
    base[ch] = val;
    ++val;
  }
  for (char ch = 'a'; ch <= 'z'; ch++) {
    base[ch] = val;
    ++val;
  }
  base['-'] = 62;
  base['_'] = 63;
  string str;
  cin >> str;
  long long cnt = 0LL;
  for (int i = 0; i < str.length(); i++) {
    int num = base[str[i]];
    bitset<6> B(num);
    B.flip();
    cnt += (long long)B.count();
  }
  printf("%lld\n", modPow(3LL, cnt, 1000000007LL));
  return 0;
}

```