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 N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
int n;
long long dp[111];
string s;
long long res = 1;
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  for (long long ii = 0; ii <= 63; ++ii) {
    for (long long i = 0; i <= 63; ++i) {
      for (long long j = 0; j <= 63; ++j) {
        if ((i & j) == ii) dp[ii]++;
      }
    }
  }
  cin >> s;
  for (long long i = 0; i <= (int)(s.size()) - 1; ++i) {
    if ('a' <= s[i] && s[i] <= 'z') {
      res *= dp[s[i] - 97 + 36];
    } else if ('A' <= s[i] && s[i] <= 'Z') {
      res *= dp[s[i] - 65 + 10];
    } else if (s[i] == '-') {
      res *= dp[62];
    } else if (s[i] == '_') {
      res *= dp[63];
    } else {
      res *= dp[s[i] - '0'];
    }
    res %= mod;
  }
  cout << res;
  return 0;
}