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;
long long int freq[1000];
char s[100005];
void initialize() {
  int i, j;
  for (i = 0; i <= 63; i++) {
    for (j = 0; j <= 63; j++) freq[(i & j)]++;
  }
}
int main() {
  scanf("%s", &s);
  initialize();
  long long int sum = 1, ad;
  for (int i = 0; i < strlen(s); i++) {
    if (s[i] == '-') ad = 62;
    if (s[i] == '_') ad = 63;
    if (('A' <= s[i]) && (s[i] <= 'Z')) ad = s[i] - 'A' + 10;
    if (('0' <= s[i]) && (s[i] <= '9')) ad = s[i] - '0';
    if (('a' <= s[i]) && (s[i] <= 'z')) ad = s[i] - 'a' + 36;
    sum = (sum * freq[ad]) % (1000000007);
  }
  printf("%I64d\n", sum);
}