Wonderful Fours

Time limit1sMemory limit128 MB

Summary

Given five digits, count unordered triples of distinct permutations (no leading zero) whose sum is a different valid permutation of the same digits.

Level: Medium5 of 10
Topics: Brute force, Combinatorics, Math, Implementation
Solved: No attempts yet

Problem

Let us call a set of five decimal digits $K_5$ . (Note that a digit may appear more than once in this set.)

We say that a five-digit natural number is properly formed from $K_5$ if it is obtained by writing all of the digits of $K_5$ in a row in some order (using each of them exactly once) and it does not start with $0$ .

For example, if $K_5$ contains the digits $1, 1, 7, 0, 4$ , then $17140$ and $47011$ are properly formed from $K_5$ , while $17740$ is not.

We call four five-digit natural numbers $s_1, s_2, s_3, s_4$ a wonderful four of $K_5$ if all of the following hold:

$s_1$ is properly formed from $K_5$ .
$s_2$ is properly formed from $K_5$ .
$s_3$ is properly formed from $K_5$ .
$s_4$ is properly formed from $K_5$ .
$s_1, s_2, s_3, s_4$ are all different numbers.
$s_1 + s_2 + s_3 = s_4$

Count how many different wonderful fours can be formed from the five digits of the set $K_5$ given in the input. (Reordering the numbers within a wonderful four does not create a new wonderful four.)

Input

A single line containing five decimal digits separated by spaces. Each digit is between $0$ and $9$ inclusive; together the five digits form the set $K_5$ .

Output

Print, on a single line, the number of different wonderful fours that can be formed from $K_5$ .

Examples2

Example 1
Input
```
0 2 6 2 8
```
Expected output
```
4
```
Example 2
Input
```
9 2 3 3 9
```
Expected output
```
0
```