#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
long long base[6 * N];
int main() {
  ios::sync_with_stdio(NULL);
  base[0] = 1;
  for (int i = 1; i < 6 * N; ++i) {
    base[i] = base[i - 1] * 3 % MOD;
  }
  string word;
  cin >> word;
  int cnt = 0;
  for (int i = 0; i < word.size(); ++i) {
    int idx;
    if (word[i] >= 'A' && word[i] <= 'Z') {
      idx = word[i] - 'A' + 10;
    } else if (word[i] >= 'a' && word[i] <= 'z') {
      idx = word[i] - 'a' + 36;
    } else if (word[i] >= '0' && word[i] <= '9') {
      idx = word[i] - '0';
    } else if (word[i] == '-') {
      idx = 62;
    } else {
      idx = 63;
    }
    for (int j = 0; j < 6; ++j) {
      if (!(idx & (1 << j))) {
        ++cnt;
      }
    }
  }
  long long ans = 1;
  ans = ans * base[cnt] % MOD;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 1e5 + 10;
const int MOD = 1e9 + 7;
long long base[6 * N];
int main() {
  ios::sync_with_stdio(NULL);
  base[0] = 1;
  for (int i = 1; i < 6 * N; ++i) {
    base[i] = base[i - 1] * 3 % MOD;
  }
  string word;
  cin >> word;
  int cnt = 0;
  for (int i = 0; i < word.size(); ++i) {
    int idx;
    if (word[i] >= 'A' && word[i] <= 'Z') {
      idx = word[i] - 'A' + 10;
    } else if (word[i] >= 'a' && word[i] <= 'z') {
      idx = word[i] - 'a' + 36;
    } else if (word[i] >= '0' && word[i] <= '9') {
      idx = word[i] - '0';
    } else if (word[i] == '-') {
      idx = 62;
    } else {
      idx = 63;
    }
    for (int j = 0; j < 6; ++j) {
      if (!(idx & (1 << j))) {
        ++cnt;
      }
    }
  }
  long long ans = 1;
  ans = ans * base[cnt] % MOD;
  printf("%I64d\n", ans);
  return 0;
}

```