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;
long long MOD = 1000000007;
string s;
long long t[10000];
void stor() {
  int i, j, cnt;
  for (i = 0; i < 64; i++) {
    j = i;
    cnt = 0;
    while (j) {
      if (j % 2 == 1) {
        cnt++;
      }
      j = j / 2;
    }
    t[i] = cnt;
  }
}
int main() {
  stor();
  long long l, i, ss = 0, a, c;
  cin >> s;
  l = s.size();
  for (i = 0; i < l; i++) {
    if ((s[i] >= '0') && (s[i] <= '9')) {
      ss += 6 - t[s[i] - '0'];
    } else if ((s[i] >= 'A') && (s[i] <= 'Z')) {
      ss += 6 - t[s[i] - 'A' + 10];
    } else if ((s[i] >= 'a') && (s[i] <= 'z')) {
      ss += 6 - t[s[i] - 'a' + 36];
    } else if (s[i] == '-') {
      ss += 6 - t[62];
    } else if (s[i] == '_') {
      ss += 6 - t[63];
    }
  }
  a = 3;
  c = 1;
  while (ss) {
    if (ss % 2 == 1) {
      c = c * a;
      c = c % MOD;
    }
    ss = ss / 2;
    a = a * a;
    a = a % MOD;
  }
  cout << c << endl;
  return 0;
}
