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;
const long long mod = 1000000007;
long long n;
string st;
long long pow(long long a, long long b) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) ret *= a;
    b /= 2;
    a *= a;
  }
  return ret;
}
long long aton(char c) {
  if (c == '-') return 62;
  if (c == '_') return 63;
  if (c <= '9' && c >= '0') return c - '0';
  if (c <= 'Z' && c >= 'A') return (c - 'A') + 10;
  if (c <= 'z' && c >= 'a') return (c - 'a') + 36;
}
long long pos(long long xx) {
  long long cnt = 0;
  for (long long i = 0; i < 6; i++) {
    if (!((1 << i) & (xx))) cnt++;
  }
  return pow(3, cnt);
}
int main() {
  long long cval, ans = 1;
  cin >> st;
  n = st.length();
  for (long long i = 0; i < n; i++) {
    cval = aton(st[i]);
    ans *= pos(cval);
    if (ans >= mod) ans %= mod;
  }
  cout << ans << endl;
}