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>
int dx[] = {0, 1, 0, -1, 1, -1, -1, 1};
int dy[] = {1, 0, -1, 0, 1, 1, -1, -1};
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int val(char c) {
  if (c >= '0' && c <= '9')
    return (int)(c - '0');
  else if (c >= 'A' && c <= 'Z')
    return (int)(c - 'A' + 10);
  else if (c >= 'a' && c <= 'z')
    return (int)(c - 'a' + 36);
  else if (c == '-')
    return 62;
  else
    return 63;
}
using namespace std;
long long int ans = 1;
int MOD = 1000000000 + 7;
void binary(int x) {
  int i = 0;
  while (x != 0) {
    i++;
    int y = x % 2;
    x /= 2;
    if (y == 0) ans *= 3;
    ans %= MOD;
  }
  while (i != 6) {
    ans *= 3;
    ans %= MOD;
    i++;
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int i, j, k, it, t;
  {
    string s;
    cin >> s;
    ans = 1;
    for (i = 0; i < s.size(); i++) {
      int tmp = val(s[i]);
      binary(tmp);
    }
    cout << ans << endl;
  }
  return 0;
}
