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>
const int maxn = 2e5 + 5;
const long long mod = 1e9 + 7;
using namespace std;
char s[maxn];
long long ci(long long x, long long t) {
  long long res = 1;
  while (t) {
    if (t & 1) res *= x;
    t >>= 1;
    x *= x;
  }
  return res;
}
long long dd(int x) {
  long long res = 0;
  while (x) {
    if (x % 2) res++;
    x /= 2;
  }
  return 6 - res;
}
long long ans = 1;
int main() {
  scanf("%s", s + 1);
  int len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    if (s[i] >= '0' && s[i] <= '9')
      ans *= ci(3LL, dd(s[i] - '0')), ans %= mod;
    else if (s[i] >= 'A' && s[i] <= 'Z')
      ans *= ci(3LL, dd(s[i] - 'A' + 10)), ans %= mod;
    else if (s[i] >= 'a' && s[i] <= 'z')
      ans *= ci(3LL, dd(s[i] - 'a' + 36)), ans %= mod;
    else if (s[i] == '-')
      ans *= 3LL, ans %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}
