#include <bits/stdc++.h>
using namespace std;
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long solve(long long x) {
  long long n = x;
  long long z = 0;
  while (n > 0) {
    n &= n - 1;
    z++;
  }
  return (6 - z);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  map<char, long long> ascii;
  for (long long i = 0; i < 26; i++) {
    ascii['a' + i] = 36 + i;
  }
  for (long long i = 0; i < 26; i++) {
    ascii['A' + i] = 10 + i;
  }
  for (long long i = 0; i < 10; i++) {
    ascii['0' + i] = 0 + i;
  }
  ascii['-'] = 62;
  ascii['_'] = 63;
  long long ans = 1;
  string s;
  cin >> s;
  for (long long i = 0; i < s.length(); i++) {
    ans = mul(ans, powm(3, solve(ascii[s[i]])));
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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 powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long solve(long long x) {
  long long n = x;
  long long z = 0;
  while (n > 0) {
    n &= n - 1;
    z++;
  }
  return (6 - z);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  map<char, long long> ascii;
  for (long long i = 0; i < 26; i++) {
    ascii['a' + i] = 36 + i;
  }
  for (long long i = 0; i < 26; i++) {
    ascii['A' + i] = 10 + i;
  }
  for (long long i = 0; i < 10; i++) {
    ascii['0' + i] = 0 + i;
  }
  ascii['-'] = 62;
  ascii['_'] = 63;
  long long ans = 1;
  string s;
  cin >> s;
  for (long long i = 0; i < s.length(); i++) {
    ans = mul(ans, powm(3, solve(ascii[s[i]])));
  }
  cout << ans;
  return 0;
}

```