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;
bool qf = false;
map<char, long long> a;
long long i, j;
long long power(long long a, long long b) {
  long long i, p = 1;
  for (i = 1; i <= b; i++) p *= a;
  return p;
}
long long len, ans = 1, cnt, num, t;
char c[100001];
int main() {
  for (i = 0; i < 10; i++) a[i + '0'] = i;
  for (i = 0; i < 26; i++) a[i + 'A'] = i + 10;
  for (i = 0; i < 26; i++) a[i + 'a'] = i + 36;
  a['-'] = 62;
  a['_'] = 63;
  scanf(" %s", c);
  len = strlen(c);
  for (i = 0; i < len; i++) {
    cnt = t = 0;
    num = a[c[i]];
    while (cnt < 6) t += (num % 2 == 0 ? 1 : 0), num /= 2, cnt++;
    ans *= power(3, t);
    ans %= 1000000007;
  }
  printf("%lld", ans);
  return 0;
}