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(int, char**) {
  string str;
  cin >> str;
  vector<int> data;
  for (auto c : str) {
    if (c >= '0' && c <= '9') {
      data.push_back(int(c - '0'));
    } else if (c >= 'A' && c <= 'Z') {
      data.push_back(int(c - 'A') + 10);
    } else if (c >= 'a' && c <= 'z') {
      data.push_back(int(c - 'a') + 36);
    } else if (c == '_') {
      data.push_back(63);
    } else if (c == '-') {
      data.push_back(62);
    }
  }
  vector<unsigned long long> map(64, 0);
  for (int i = 0; i < 64; ++i)
    for (int j = 0; j < 64; ++j) {
      map[i & j] += 1;
    }
  unsigned long long res = 1;
  unsigned long long magicN = 1e9 + 7;
  for (auto d : data) {
    unsigned long long a = (res % magicN);
    unsigned long long b = (map[d] % magicN);
    res = a * b;
  }
  res %= magicN;
  cout << res;
  return 0;
}
