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;
int getz(char c) {
  if (c >= '0' && c <= '9') {
    return c - '0';
  }
  if (c >= 'A' && c <= 'Z') {
    return 10 + (c - 'A');
  }
  if (c >= 'a' && c <= 'z') {
    return 36 + (c - 'a');
  }
  if (c == '-') {
    return 62;
  }
  return 63;
}
vector<int> get2(int n) {
  vector<int> ans(6, 0);
  int i = 0;
  while (n) {
    ans[i] = (n % 2);
    ++i;
    n /= 2;
  }
  return ans;
}
long long pw3(long long p) {
  if (!p) {
    return 1;
  }
  if (!(p % 2)) {
    long long ans = pw3(p / 2);
    ans *= ans;
    ans %= (long long)(1e9 + 7);
    return ans;
  } else {
    long long ans = pw3(p - 1);
    ans *= 3;
    ans %= (long long)(1e9 + 7);
    return ans;
  }
}
void solve() {
  string s;
  cin >> s;
  long long st3 = 0;
  for (int(i) = 0; (i) < (s.size()); ++(i)) {
    int x = getz(s[i]);
    vector<int> mf = get2(x);
    for (int(j) = 0; (j) < (mf.size()); ++(j)) {
      if (!mf[j]) {
        ++st3;
      }
    }
  }
  cout << pw3(st3);
}
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}