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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  map<char, int> base;
  char ch;
  ch = '0';
  for (long long int i = 0; i < 10; i++) {
    base[ch] = i;
    ch++;
  }
  ch = 'A';
  for (long long int i = 10; i < 36; i++) {
    base[ch] = i;
    ch++;
  }
  ch = 'a';
  for (long long int i = 36; i < 62; i++) {
    base[ch] = i;
    ch++;
  }
  ch = '-';
  base[ch] = 62;
  ch = '_';
  base[ch] = 63;
  long long int ways[64];
  long long int rem, store;
  for (long long int i = 0; i < 64; i++) {
    long long int temp = 1;
    store = i;
    for (long long int j = 1; j <= 6; j++) {
      rem = store % 2;
      store = store / 2;
      if (rem == 0) {
        temp = temp * 3;
      }
    }
    ways[i] = temp;
  }
  long long int ans = 1, len, val;
  string str;
  cin >> str;
  len = str.size();
  for (long long int i = 0; i < len; i++) {
    val = ways[base[str[i]]];
    ans = (ans * val) % 1000000007;
  }
  cout << ans;
  return 0;
}
