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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define reps(i,start,end,step) for(int i=start;i< end;i+=step)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
#define Pr(x) cout << x << endl
#define Prs(x) cout << x << " "



int main()
{
    ll n1, n2, N, M;
    cin >> n1 >> n2;
    if(n1 <= n2){
        N = n1;
        M = n2;
    }
    else{
        N = n2;
        M = n1;
    }
            
    ll ans =0;
    if ((N==1)&(M==1)) ans =1;
    else if((N==1)& (M>=2)) ans = M-2;
    else if((N==2)&(M==2)) ans = 0;
    else if((N==2)&(M>=3)) ans = 0;
    else if((N>=3)&(M>=3)) ans = (N-2) * (M-2);
    cout << ans  << endl;
    
    return 0;
}