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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const long long mod = 1e9 + 7;
char s[100100];
int get(char s) {
  if (s >= '0' && s <= '9') return s - '0';
  if (s >= 'a' && s <= 'z') return s - 'a' + 36;
  if (s >= 'A' && s <= 'Z') return s - 'A' + 10;
  if (s == '-') return 62;
  if (s == '_') return 63;
}
int main() {
  scanf("%s", s);
  int len = strlen(s);
  long long ans1 = 0, ans2 = 0;
  long long ans = 1;
  bool ok = false;
  for (int i = 0; i < len; i++) {
    int temp = get(s[i]);
    for (int j = 5; j >= 0; j--) {
      if ((temp >> j) & 1) {
        ok = true;
      } else
        ans = (ans * 3) % mod;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}