#include <bits/stdc++.h>
using namespace std;
long long int getZeroes(int A) {
  long long int ans = 0;
  for (int i = 0; i < 6; i++) {
    if (!(A & (1 << i))) {
      ans++;
    }
  }
  return ans;
}
long long int modMul(long long int A, long long int B) {
  return (A * B) % 1000000007;
}
long long int modAdd(long long int A, long long int B) {
  return (A + B) % 1000000007;
}
int getVal(char A) {
  if (A >= '0' && A <= '9') {
    return A - '0';
  }
  if (A >= 'A' && A <= 'Z') {
    return (A - 'A') + 10;
  }
  if (A >= 'a' && A <= 'z') {
    return (A - 'a') + 36;
  }
  if (A == '-') {
    return 62;
  }
  if (A == '_') {
    return 63;
  }
}
string S = "";
long long int finalAns = 1;
int main() {
  cin >> S;
  for (int i = 0; i < S.size(); i++) {
    long long int val = getZeroes(getVal(S[i]));
    long long int X = 1;
    while (val) {
      X = modMul(3ll, X);
      val--;
    }
    finalAns = modMul(finalAns, X);
  }
  finalAns %= 1000000007;
  printf("%lld\n", finalAns);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int getZeroes(int A) {
  long long int ans = 0;
  for (int i = 0; i < 6; i++) {
    if (!(A & (1 << i))) {
      ans++;
    }
  }
  return ans;
}
long long int modMul(long long int A, long long int B) {
  return (A * B) % 1000000007;
}
long long int modAdd(long long int A, long long int B) {
  return (A + B) % 1000000007;
}
int getVal(char A) {
  if (A >= '0' && A <= '9') {
    return A - '0';
  }
  if (A >= 'A' && A <= 'Z') {
    return (A - 'A') + 10;
  }
  if (A >= 'a' && A <= 'z') {
    return (A - 'a') + 36;
  }
  if (A == '-') {
    return 62;
  }
  if (A == '_') {
    return 63;
  }
}
string S = "";
long long int finalAns = 1;
int main() {
  cin >> S;
  for (int i = 0; i < S.size(); i++) {
    long long int val = getZeroes(getVal(S[i]));
    long long int X = 1;
    while (val) {
      X = modMul(3ll, X);
      val--;
    }
    finalAns = modMul(finalAns, X);
  }
  finalAns %= 1000000007;
  printf("%lld\n", finalAns);
  return 0;
}

```