#include <bits/stdc++.h>
using namespace std;
long long noOfUnsetBits(char c) {
  long long num;
  if (c >= '0' && c <= '9') {
    num = c - '0';
  } else if (c >= 'A' && c <= 'Z') {
    num = 10 + (c - 'A');
  } else if (c >= 'a' && c <= 'z') {
    num = 36 + (c - 'a');
  } else if (c == '-') {
    num = 62;
  } else if (c == '_') {
    num = 63;
  } else {
    assert(false);
  }
  long long unsetBits = 0;
  for (long long i = 0; i < 6; i++) {
    if ((num >> i) & 1) {
    } else {
      unsetBits++;
    }
  }
  return unsetBits;
}
const long long MOD = 1e9 + 7;
long long powerModulo(long long base, long long power) {
  long long res = 1;
  while (power) {
    if (power & 1) res = (res * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return res;
}
int32_t main() {
  string s;
  cin >> s;
  long long total_unset = 0;
  for (char c : s) total_unset += noOfUnsetBits(c);
  cout << powerModulo(3ll, total_unset) << "\n";
  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 noOfUnsetBits(char c) {
  long long num;
  if (c >= '0' && c <= '9') {
    num = c - '0';
  } else if (c >= 'A' && c <= 'Z') {
    num = 10 + (c - 'A');
  } else if (c >= 'a' && c <= 'z') {
    num = 36 + (c - 'a');
  } else if (c == '-') {
    num = 62;
  } else if (c == '_') {
    num = 63;
  } else {
    assert(false);
  }
  long long unsetBits = 0;
  for (long long i = 0; i < 6; i++) {
    if ((num >> i) & 1) {
    } else {
      unsetBits++;
    }
  }
  return unsetBits;
}
const long long MOD = 1e9 + 7;
long long powerModulo(long long base, long long power) {
  long long res = 1;
  while (power) {
    if (power & 1) res = (res * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return res;
}
int32_t main() {
  string s;
  cin >> s;
  long long total_unset = 0;
  for (char c : s) total_unset += noOfUnsetBits(c);
  cout << powerModulo(3ll, total_unset) << "\n";
  return 0;
}

```