Problem:
Score : 300 points

Problem Statement

There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region.
The front and back sides of these cards can be distinguished, and initially every card faces up.

We will perform the following operation once for each square contains a card:

For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square.

It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed.
Find the number of cards that face down after all the operations.

Constraints

1 \leq N,M \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of cards that face down after all the operations.

Sample Input 1

2 2

Sample Output 1

0

We will flip every card in any of the four operations. Thus, after all the operations, all cards face up.

Sample Input 2

1 7

Sample Output 2

5

After all the operations, all cards except at both ends face down.

Sample Input 3

314 1592

Sample Output 3

496080

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// include
// ------------------------------------------------
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

// func
// ------------------------------------------------
int CalcSumOfDigit(int n);      // 各桁の和を計算する。
long long getDigit(long long n);      // 数字の桁数を取得する。
string upper(string str);       // 英字を大文字に変換する。
string lower(string str);       // 英字を小文字に変換する。

// define
// ------------------------------------------------
#define all(a)        (a).begin(),(a).end()
#define rall(a)       (a).rbegin(),(a).rend()
#define sz(a)         int((a).size())
#define rep(i,n)      for(int(i)=0;(i)<(n);(i)++)
#define repe(i,n)     for(int(i)=0;(i)<=(n);(i)++)
#define vsort(v)      sort((v).begin(),(v).end())
#define rvsort(v)     sort(rall((v)))
#define vi            vector<int>
#define GCD(a,b)      __gcd((a),(b))
#define LCM(a,b)      ((a)/GCD((a),(b))*(b))
#define kiriage(a,b)  ((a)+(b)-1)/(b)

const int INF = 1e9;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vll;

// code
// ------------------------------------------------
int main()
{
  ll n,m;
  cin >> n >> m;

  ll ans;
  if(n == 1 && m == 1)
  {
    ans = 1;
  }
  else if(n == 1)
  {
    ans = m - 2;
  }
  else if (m == 1)
  {
    ans = n - 2;
  }
  else if(n == 2 || m == 2)
  {
    ans = 0;
  }
  else
  {
    ans = (n - 2) * ( m - 2);
  }
  
  cout << ans << endl;


  return 0;
}
// funcの実体
// ------------------------------------------------
ll getDigit(ll n)
{
  int i = 1;
  while(1)
  {
    n = n / 10;
    if(n == 0)
      break;
    i++;
  }
  return i;
}
ll CalcSumOfDigit(ll n)
{
  ll s = 0;
  while(n)
  {
    s += n % 10;
    n = n / 10;
  }
  return s;
}

string upper(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(97 <= *itr && *itr <= 122)
    {
      *itr = *itr - 32;
    }
  }
  return str;
}

string lower(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(65 <= *itr && *itr <= 90)
    {
      *itr = *itr + 32;
    }
  }
  return str;
}