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;
inline long long input() {
  long long n;
  cin >> n;
  return n;
}
long long pw(long long a, long long b) {
  return (!b ? 1 : (b & 1 ? a * pw(a * a, b / 2) : pw(a * a, b / 2)));
}
const long long maxn = 1e5 + 10, mod = 1e9 + 7;
long long ans = 1;
long long cnt[100];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  string x;
  cin >> x;
  for (int i = 0; i < 64; i++) {
    for (int j = 0; j < 64; j++) {
      cnt[i & j]++;
    }
  }
  for (int i = 0; i < x.size(); i++) {
    int y = (int)x[i];
    if (y >= 65 && y <= 90) {
      ans *= cnt[y - 55] % mod;
      ans %= mod;
    } else if (y >= 97 && y <= 122) {
      ans *= cnt[y - 61] % mod;
      ans %= mod;
    } else if (y >= 48 && y <= 57) {
      ans *= cnt[y - 48] % mod;
      ans %= mod;
    } else if (y == 45) {
      ans *= cnt[62] % mod;
      ans %= mod;
    } else if (y == 95) {
      ans *= cnt[63] % mod;
      ans %= mod;
    }
  }
  ans %= mod;
  cout << ans % mod;
  return 0;
}