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;
const long long mod = 1000000007LL;
map<char, int> mp;
char in[1000005];
long long pow_mod(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
int f(int x) {
  x = (~x) & ((1 << 6) - 1);
  int ans = 0;
  while (x) ans += x & 1, x >>= 1;
  return ans;
}
int main() {
  for (int i = 0; i <= 9; ++i) mp[i + '0'] = i;
  for (int i = 'A'; i <= 'Z'; ++i) mp[i] = i - 'A' + 10;
  for (int i = 'a'; i <= 'z'; ++i) mp[i] = i - 'a' + 36;
  mp['-'] = 62;
  mp['_'] = 63;
  int cnt = 0;
  gets(in);
  int x;
  for (char* p = in; *p; ++p) {
    x = mp[*p];
    cnt += f(x);
  }
  printf("%I64d\n", pow_mod(3, cnt));
  return 0;
}
