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;
string s;
int a[64], cou;
long long int Mod = 1e9 + 7;
void compute() {
  for (int i = 0; i < 64; i++) {
    int n = i, sum = 6;
    while (n != 0) {
      while (n % 2 == 0) n /= 2;
      sum--;
      n = n / 2;
    }
    a[i] = sum;
  }
  return;
}
void fast(int num) {
  long long int k = 3, ans = 1;
  while (num != 0) {
    if (num % 2 == 0) {
      k *= k;
      num /= 2;
      k %= Mod;
    } else
      ans *= k, num--;
    if (ans > Mod) ans %= Mod;
  }
  cout << ans << endl;
}
int main() {
  compute();
  cin >> s;
  cou = 0;
  for (int x = 0; x < s.length(); x++) {
    if (s[x] >= '0' && s[x] <= '9')
      s[x] = s[x] - '0';
    else if (s[x] >= 'a' && s[x] <= 'z')
      s[x] = s[x] - 'a' + 36;
    else if (s[x] >= 'A' && s[x] <= 'Z')
      s[x] = s[x] - 'A' + 10;
    else if (s[x] == '-')
      s[x] = 62;
    else if (s[x] == '_')
      s[x] = 63;
    cou += a[s[x]];
  }
  fast(cou);
  return 0;
}
