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;
void Egriiiiiiiii() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
unsigned int countSetBits(unsigned int n) {
  unsigned int count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
int mod = 1e9 + 7;
int fpow(int b, int p) {
  if (!p) return 1;
  int ret = fpow(b, p / 2) % mod;
  return (p & 1 ? (((ret % mod) * (ret % mod) * (b % mod)) % mod)
                : (((ret % mod) * (ret % mod)) % mod));
}
map<long long, bool> mp;
int main() {
  Egriiiiiiiii();
  string s;
  cin >> s;
  int n;
  long long ans = 1, t;
  for (int i = 0; i < (int)s.size(); i++) {
    t = 0;
    if (s[i] == '-')
      n = 62;
    else if (s[i] == '_')
      n = 63;
    else if (s[i] >= 'a' && s[i] <= 'z')
      n = (s[i] - 'a') + 36;
    else if (s[i] >= '0' && s[i] <= '9')
      n = s[i] - '0';
    else
      n = (s[i] - 'A') + 10;
    while (n) {
      if (n % 2 == 1) t++;
      n /= 2;
    }
    t = fpow(1LL * 3, 6 - t);
    t %= mod;
    ans = (ans % mod) * t;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}