#include <bits/stdc++.h>
long long int mod = 1000000007;
using namespace std;
void fillDivisior(vector<long long int>& divisior, long long int x) {
  for (long long int y = 1; y <= sqrt(x); y++) {
    if (x % y == 0) {
      divisior.push_back(y);
      ;
      if (x / y != y) divisior.push_back(x / y);
      ;
    }
  }
}
int main() {
  long long int n;
  int arr[64];
  for (int i = 0; i < 64; i++) {
    arr[i] = 0;
  }
  for (int i = 0; i < 64; i++) {
    for (int j = 0; j < 64; j++) {
      arr[i & j]++;
    }
  }
  string ss;
  cin >> ss;
  long long int ans = 1;
  for (int i = 0; i < ss.length(); i++) {
    int temp;
    if (ss[i] == '_') {
      temp = 63;
    } else if (ss[i] == '-') {
      temp = 62;
    } else if (ss[i] >= 'a' && ss[i] <= 'z') {
      temp = (ss[i] - 'a') + 36;
    } else if (ss[i] >= 'A' && ss[i] <= 'Z') {
      temp = (ss[i] - 'A') + 10;
    } else {
      temp = ss[i] - '0';
    }
    ans = (ans * (arr[temp])) % mod;
  }
  cout << ans % mod;
}

### Prompt

Please create a solution in cpp 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>
long long int mod = 1000000007;
using namespace std;
void fillDivisior(vector<long long int>& divisior, long long int x) {
  for (long long int y = 1; y <= sqrt(x); y++) {
    if (x % y == 0) {
      divisior.push_back(y);
      ;
      if (x / y != y) divisior.push_back(x / y);
      ;
    }
  }
}
int main() {
  long long int n;
  int arr[64];
  for (int i = 0; i < 64; i++) {
    arr[i] = 0;
  }
  for (int i = 0; i < 64; i++) {
    for (int j = 0; j < 64; j++) {
      arr[i & j]++;
    }
  }
  string ss;
  cin >> ss;
  long long int ans = 1;
  for (int i = 0; i < ss.length(); i++) {
    int temp;
    if (ss[i] == '_') {
      temp = 63;
    } else if (ss[i] == '-') {
      temp = 62;
    } else if (ss[i] >= 'a' && ss[i] <= 'z') {
      temp = (ss[i] - 'a') + 36;
    } else if (ss[i] >= 'A' && ss[i] <= 'Z') {
      temp = (ss[i] - 'A') + 10;
    } else {
      temp = ss[i] - '0';
    }
    ans = (ans * (arr[temp])) % mod;
  }
  cout << ans % mod;
}

```