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;
map<char, int> mp;
int vis[10010];
string s;
int main() {
  char c;
  for (int i = 0; i <= 9; i++) {
    c = i + '0';
    mp[c] = i;
  }
  int num = 0;
  for (int i = 10; i <= 35; i++) {
    c = num + 'A';
    mp[c] = i;
    num++;
  }
  num = 0;
  for (int i = 36; i <= 61; i++) {
    c = num + 'a';
    mp[c] = i;
    num++;
  }
  c = '-';
  mp[c] = 62;
  c = '_';
  mp[c] = 63;
  map<char, int>::iterator it;
  map<char, int>::iterator it1;
  for (it = mp.begin(); it != mp.end(); it++) {
    int x = it->second;
    for (it1 = mp.begin(); it1 != mp.end(); it1++) {
      int y = it1->second;
      vis[x & y]++;
    }
  }
  long long ans = 1;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    char c = s[i];
    ans = ans * vis[mp[c]] % 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}