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 MAXINT = 2147483640;
const long long MAXLL = 9223372036854775800LL;
const long long MAXN = 1e6;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
using namespace std;
long long f[MAXN];
string s;
long long bp(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a, res %= mod;
    a *= a, a %= mod;
    n /= 2;
  }
  return res;
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  for (int i = '0'; i <= '9'; ++i) f[i] = i - '0';
  for (int i = 'A'; i <= 'Z'; ++i) f[i] = i - 'A' + 10;
  for (int i = 'a'; i <= 'z'; ++i) f[i] = i - 'a' + 36;
  f['-'] = 62;
  f['_'] = 63;
  cin >> s;
  long long ans = 1;
  for (int i = 0; i < s.size(); ++i) {
    long long x = f[s[i]];
    long long kol = 0;
    for (int x1 = 0; x1 <= 63; ++x1)
      for (int x2 = 0; x2 <= 63; ++x2)
        if ((x1 & x2) == x) kol++;
    ans *= kol;
    ans %= mod;
  }
  cout << ans << "\n";
  return 0;
}
