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 inf = 1e18;
const long double eps = 1e-6;
const long long mod = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  string all;
  for (char i = '0'; i <= '9'; ++i) all.push_back(i);
  for (char i = 'A'; i <= 'Z'; ++i) all.push_back(i);
  for (char i = 'a'; i <= 'z'; ++i) all.push_back(i);
  all.push_back('-');
  all.push_back('_');
  map<char, long long> idx;
  for (long long i = 0; i < (long long)all.size(); ++i) idx[all[i]] = i;
  long long ans = 1;
  string s;
  cin >> s;
  for (char c : s) {
    long long x = idx[c];
    for (long long i = 0; i < 6; ++i) {
      if (!((1 << i) & x)) ans = (ans * 3) % mod;
    }
  }
  cout << ans;
  return 0;
}