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
#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
struct node {
  int x;
  int y;
} b[N];
bool cmp(struct node a, struct node b) {
  if (a.x < b.x) {
    return true;
  }
  return false;
}
int main() {
  char a[N];
  while (cin >> a) {
    int num;
    long long int ans = 1;
    for (int i = 0; i < strlen(a); i++) {
      if (a[i] >= '0' && a[i] <= '9') {
        num = a[i] - '0';
      } else if (a[i] >= 'A' && a[i] <= 'Z') {
        num = a[i] - 'A' + 10;
      } else if (a[i] >= 'a' && a[i] <= 'z') {
        num = a[i] - 'a' + 36;
      } else if (a[i] == '-') {
        num = 62;
      } else {
        num = 63;
      }
      for (int j = 0; j <= 5; j++) {
        if (((num >> j) & 1) == 0) {
          ans = ans * 3 % 1000000007;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
