#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int mod = 1e9 + 7;
const int maxn = 100000 + 5;
int T, n, m;
int a[100];
char s[maxn];
int main() {
  memset(a, 0, sizeof(a));
  for (int k = 0; k < 64; k++) {
    for (int i = 0; i < 64; i++)
      for (int j = 0; j < 64; j++) {
        if ((i & j) == k) {
          if (i != j)
            a[k]++;
          else
            a[k]++;
        }
      }
  }
  cin >> s;
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    if (s[i] >= '0' && s[i] <= '9')
      s[i] -= '0';
    else if (s[i] >= 'A' && s[i] <= 'Z')
      s[i] += 10 - 'A';
    else if (s[i] >= 'a' && s[i] <= 'z')
      s[i] += 36 - 'a';
    else if (s[i] == '-')
      s[i] = 62;
    else if (s[i] == '_')
      s[i] = 63;
  }
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    ans = ans * a[s[i]] % mod;
  }
  cout << ans << endl;
  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 double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int mod = 1e9 + 7;
const int maxn = 100000 + 5;
int T, n, m;
int a[100];
char s[maxn];
int main() {
  memset(a, 0, sizeof(a));
  for (int k = 0; k < 64; k++) {
    for (int i = 0; i < 64; i++)
      for (int j = 0; j < 64; j++) {
        if ((i & j) == k) {
          if (i != j)
            a[k]++;
          else
            a[k]++;
        }
      }
  }
  cin >> s;
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    if (s[i] >= '0' && s[i] <= '9')
      s[i] -= '0';
    else if (s[i] >= 'A' && s[i] <= 'Z')
      s[i] += 10 - 'A';
    else if (s[i] >= 'a' && s[i] <= 'z')
      s[i] += 36 - 'a';
    else if (s[i] == '-')
      s[i] = 62;
    else if (s[i] == '_')
      s[i] = 63;
  }
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    ans = ans * a[s[i]] % mod;
  }
  cout << ans << endl;
  return 0;
}

```