#include <bits/stdc++.h>
using namespace std;
long long int logy(long long int x, long long int y) {
  return ceil((long double)(log(x) / log(y)));
}
long long int powe(long long int a, long long int b) {
  long long int re = 1;
  while (b) {
    if (b & 1) re = (re * a) % 1000000007;
    a = (a * a) % 1000000007;
    b = b >> 1;
  }
  return re;
}
bool flag = 0, flag1 = 0, flag2 = 0;
long long int get(long long int n) {
  long long int cnt = 0;
  for (long long int i = 0; i < 6; i++)
    if ((n & 1 << i) == 0) cnt++;
  return powe(3, cnt);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1;
  while (tt--) {
    string s;
    cin >> s;
    long long int ans[s.length()];
    for (long long int i = 0; i < s.length(); i++) {
      if (s[i] - '0' >= 0 && s[i] - '0' <= 9)
        ans[i] = (long long int)(s[i] - '0');
      else if (s[i] >= 65 && s[i] <= 90)
        ans[i] = (long long int)(s[i] - 'A' + 10);
      else if (s[i] >= 97 && s[i] <= 122)
        ans[i] = (long long int)(s[i] - 'a' + 36);
      else if (s[i] == '-')
        ans[i] = 62;
      else
        ans[i] = 63;
    }
    long long int fans = 1;
    for (long long int i = 0; i < s.length(); i++) {
      fans = (fans * get(ans[i])) % 1000000007;
    }
    cout << fans;
  }
  return 0;
}

### 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>
using namespace std;
long long int logy(long long int x, long long int y) {
  return ceil((long double)(log(x) / log(y)));
}
long long int powe(long long int a, long long int b) {
  long long int re = 1;
  while (b) {
    if (b & 1) re = (re * a) % 1000000007;
    a = (a * a) % 1000000007;
    b = b >> 1;
  }
  return re;
}
bool flag = 0, flag1 = 0, flag2 = 0;
long long int get(long long int n) {
  long long int cnt = 0;
  for (long long int i = 0; i < 6; i++)
    if ((n & 1 << i) == 0) cnt++;
  return powe(3, cnt);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1;
  while (tt--) {
    string s;
    cin >> s;
    long long int ans[s.length()];
    for (long long int i = 0; i < s.length(); i++) {
      if (s[i] - '0' >= 0 && s[i] - '0' <= 9)
        ans[i] = (long long int)(s[i] - '0');
      else if (s[i] >= 65 && s[i] <= 90)
        ans[i] = (long long int)(s[i] - 'A' + 10);
      else if (s[i] >= 97 && s[i] <= 122)
        ans[i] = (long long int)(s[i] - 'a' + 36);
      else if (s[i] == '-')
        ans[i] = 62;
      else
        ans[i] = 63;
    }
    long long int fans = 1;
    for (long long int i = 0; i < s.length(); i++) {
      fans = (fans * get(ans[i])) % 1000000007;
    }
    cout << fans;
  }
  return 0;
}

```