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;
char s[111111];
long long getnum(char c) {
  if ('0' <= c && c <= '9')
    return c - '0';
  else if ('A' <= c && c <= 'Z')
    return c - 'A' + 10;
  else if ('a' <= c && c <= 'z')
    return c - 'a' + 36;
  if (c == '-')
    return 62;
  else
    return 63;
}
long long chk(char c) {
  long long ret = 0, num = getnum(c);
  for (long long i = 0; i < 6; i++)
    if (((1 << i) & num) == 0) ret++;
  return ret;
}
int main() {
  scanf("%s", s);
  long long len = strlen(s), mul = 0, fs = 0, num = getnum(s[0]), ans = 1;
  for (long long i = 0; i < len; i++) mul += chk(s[i]);
  for (long long i = 0; i < mul; i++) ans = (ans * 3LL) % 1000000007;
  printf("%I64d\n", ans);
}