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>
int chart[64];
void ini_chart() {
  int i, j, k;
  memset(chart, 0, sizeof(chart));
  for (i = 0; i < 64; i++) {
    for (j = 0; j < 6; j++) {
      k = i >> j;
      if ((k % 2) == 0) chart[i]++;
    }
  }
  return;
}
int main() {
  char data[100000];
  int j, k, i, l, mod = 1000000007;
  unsigned int ans = 1;
  ini_chart();
  scanf("%s", data);
  j = strlen(data);
  for (i = 0; i < j; i++) {
    if (data[i] == '-') {
      k = 62;
    } else if (data[i] == '_') {
      k = 63;
    } else if (data[i] < 58) {
      k = data[i] - 48;
    } else if (data[i] < 91) {
      k = data[i] - 65 + 10;
    } else {
      k = data[i] - 97 + 36;
    }
    for (l = 0; l < chart[k]; l++) {
      ans = (ans * 3) % mod;
    }
  }
  printf("%d", ans);
  return 0;
}