#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.141592653589793238462643383279502884L;
const long double E = 2.718281828459045235360287471352662498L;
const long double EPS = 1e-12L;
int64_t a[64];
int64_t r;
int n;
int64_t mod = 1000000007;
string s;
int b[100240];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(11);
  cout.setf(ios::fixed);
  fill(a, a + 64, 0);
  for (int i = 0; i < 64; ++i) {
    for (int j = 0; j < 64; ++j) {
      ++a[i & j];
    }
  }
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    if ('0' <= s[i] && s[i] <= '9') {
      b[i] = s[i] - '0';
    } else if ('A' <= s[i] && s[i] <= 'Z') {
      b[i] = 10 + (s[i] - 'A');
    } else if ('a' <= s[i] && s[i] <= 'z') {
      b[i] = 36 + (s[i] - 'a');
    } else if (s[i] == '-') {
      b[i] = 62;
    } else if (s[i] == '_') {
      b[i] = 63;
    }
  }
  r = 1;
  for (int i = 0; i < n; ++i) {
    r *= a[b[i]];
    r %= mod;
  }
  cout << r << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const long double PI = 3.141592653589793238462643383279502884L;
const long double E = 2.718281828459045235360287471352662498L;
const long double EPS = 1e-12L;
int64_t a[64];
int64_t r;
int n;
int64_t mod = 1000000007;
string s;
int b[100240];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(11);
  cout.setf(ios::fixed);
  fill(a, a + 64, 0);
  for (int i = 0; i < 64; ++i) {
    for (int j = 0; j < 64; ++j) {
      ++a[i & j];
    }
  }
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    if ('0' <= s[i] && s[i] <= '9') {
      b[i] = s[i] - '0';
    } else if ('A' <= s[i] && s[i] <= 'Z') {
      b[i] = 10 + (s[i] - 'A');
    } else if ('a' <= s[i] && s[i] <= 'z') {
      b[i] = 36 + (s[i] - 'a');
    } else if (s[i] == '-') {
      b[i] = 62;
    } else if (s[i] == '_') {
      b[i] = 63;
    }
  }
  r = 1;
  for (int i = 0; i < n; ++i) {
    r *= a[b[i]];
    r %= mod;
  }
  cout << r << endl;
  return 0;
}

```