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>
using namespace std;
long long mod = 1000000007;
int main() {
  map<int, char> itc;
  map<char, int> cti;
  vector<char> st;
  for (int i = 0; i <= 9; i++) {
    itc[i] = (i + '0');
    cti[i + '0'] = i;
    st.push_back(i + '0');
  }
  for (char a = 'A'; a <= 'Z'; a++) {
    cti[a] = a - 55;
    itc[a - 55] = a;
    st.push_back(a);
  }
  for (char a = 'a'; a <= 'z'; a++) {
    cti[a] = a - 61;
    itc[a - 61] = a;
    st.push_back(a);
  }
  itc[62] = '-';
  cti['-'] = 62;
  itc[63] = '_';
  cti['_'] = 63;
  st.push_back('-');
  st.push_back('_');
  map<char, int> cnt;
  for (int i = 0; i < st.size(); i++) {
    for (int j = 0; j < st.size(); j++) {
      int res = cti[st[i]] & cti[st[j]];
      if (res >= 0 && res < 64) {
        cnt[itc[res]]++;
      }
    }
  }
  string str;
  cin >> str;
  long long res = 1;
  for (int i = 0; i < str.size(); i++) {
    res = (res % mod * cnt[str[i]] % mod) % mod;
  }
  cout << res << endl;
  return 0;
}
