#include <bits/stdc++.h>
using namespace std;
long long pow_mod(long long x, long long n, long long mod) {
  if (n == 0) return 1;
  long long res = pow_mod(x * x % mod, n / 2, mod);
  if (n & 1) res = res * x % mod;
  return res;
}
int popcount(int n) {
  int res = 0;
  for (int i = 0; i < 6; ++i) {
    if ((n & 1) == 0) res++;
    n >>= 1;
  }
  return res;
}
char s[100010];
int main() {
  scanf("%s", s);
  int pow = 0, tmp, len = strlen(s);
  for (int i = 0; i < len; i++) {
    if (s[i] == '_')
      tmp = 63;
    else if (s[i] == '-')
      tmp = 62;
    else if (s[i] <= '9')
      tmp = s[i] - '0';
    else if (s[i] <= 'Z')
      tmp = 10 + s[i] - 'A';
    else
      tmp = 36 + s[i] - 'a';
    pow += popcount(tmp);
  }
  printf("%I64d", pow_mod(3, pow, 1e9 + 7));
  return 0;
}

### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long pow_mod(long long x, long long n, long long mod) {
  if (n == 0) return 1;
  long long res = pow_mod(x * x % mod, n / 2, mod);
  if (n & 1) res = res * x % mod;
  return res;
}
int popcount(int n) {
  int res = 0;
  for (int i = 0; i < 6; ++i) {
    if ((n & 1) == 0) res++;
    n >>= 1;
  }
  return res;
}
char s[100010];
int main() {
  scanf("%s", s);
  int pow = 0, tmp, len = strlen(s);
  for (int i = 0; i < len; i++) {
    if (s[i] == '_')
      tmp = 63;
    else if (s[i] == '-')
      tmp = 62;
    else if (s[i] <= '9')
      tmp = s[i] - '0';
    else if (s[i] <= 'Z')
      tmp = 10 + s[i] - 'A';
    else
      tmp = 36 + s[i] - 'a';
    pow += popcount(tmp);
  }
  printf("%I64d", pow_mod(3, pow, 1e9 + 7));
  return 0;
}

```