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;
long long mod = 1e9 + 7;
char str[100050];
long long ksm(long long a, long long b) {
  long long k = a, ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * k % mod;
    b = b >> 1;
    k = k * k % mod;
  }
  return ans;
}
long long Change(char x) {
  if (x <= '9' && x >= '0') return x - '0';
  if (x <= 'Z' && x >= 'A') return x - 'A' + 10;
  if (x <= 'z' && x >= 'a') return x - 'a' + 36;
  if (x == '-') return 62;
  return 63;
}
long long Count(long long x) {
  long long ans = 0;
  for (long long i = 0; i <= 5; ++i) {
    if (((x >> i) & 1) == 0) ++ans;
  }
  return ans;
}
int main() {
  scanf("%s", str);
  int n = strlen(str);
  long long tmp, ans = 1;
  for (int i = 0; i < n; ++i) {
    tmp = Count(Change(str[i]));
    ans = ans * ksm(3, tmp) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}