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 int MOD = int(1e9 + 7);
map<char, long long> m;
long long calPow(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n % 2) res = (res * a) % MOD;
    a = (a * a) % MOD;
    n >>= 1;
  }
  return res;
}
long long calBit0(long long n) {
  long long cnt = 0;
  for (int i = 0; i < 6; i++) {
    if ((n & (1LL << i)) == 0) cnt++;
  }
  return cnt;
}
string s;
int main() {
  cin >> s;
  for (char i = '0', j = 0; i <= '9'; i++, j++) m[i] = j;
  for (char i = 'A', j = 10; i <= 'Z'; i++, j++) m[i] = j;
  for (char i = 'a', j = 36; i <= 'z'; i++, j++) m[i] = j;
  m['-'] = 62, m['_'] = 63;
  int len = s.size();
  long long cnt = 0;
  for (int i = 0; i < len; i++) cnt += calBit0(m[s[i]]);
  cout << calPow(3, cnt) << endl;
  return 0;
}