**Solution Notes (Kalki Seksaria):**

This problem can be solved by recursion, and we can try all possible paths. The part of the path consisting of '(' is constructed first, and this corresponds to second = false. Once a ')' is found, second is marked as true and only ')' can be added to the path. When numopen = numclose, this path is over. An additional optimization can optionally be made. If the '(' part of the path is over, then the length of the path, if it exists, is 2*numopen. If this cannot be better then the best answer found, then there is no reason to continue exploring this path.

```
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
#define pii pair<int,int>
int N;
bool unvis[7][7];
int graph[7][7]; //-1 is border, 0 is open, 1 is close
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int ans = 0;
void calc(int r, int c, int numopen, int numclose, bool second)
{
if(numopen == numclose)
{
ans = max(ans, numopen + numclose);
return;
}
//there is no need to check solutions that we know cannot be better than the best answer that we have found so far
//this optimization is not required to solve the given test cases within 1 second
if(second && (2*numopen <= ans))
return;
unvis[r][c] = false;
for (int i = 0; i < 4; i++)
{
int r2 = r + dx[i];
int c2 = c + dy[i];
if(unvis[r2][c2])
{
if(graph[r2][c2] == 1)
calc(r2, c2, numopen, numclose + 1, true);
else if(!second)
calc(r2, c2, numopen + 1, numclose, false);
}
}
unvis[r][c] = true;
}
int main()
{
ifstream in ("hshoe.in");
ofstream out ("hshoe.out");
in >> N;
for (int i = 0; i <= N+1; i++) //adding a border around the grid makes it easier to check if a location is within the grid
{
unvis[0][i] = unvis[N+1][i] = unvis[i][0] = unvis[i][N+1] = false;
graph[0][i] = graph[N+1][i] = graph[i][0] = graph[i][N+1] = -1;
}
for (int i = 1; i <= N; i++)
{
string s;
in >> s;
for (int j = 1; j <= N; j++)
{
unvis[i][j] = true;
if(s[j-1] == '(')
graph[i][j] = 0;
else
graph[i][j] = 1;
}
}
if(graph[1][1] == 0)
calc(1,1,1,0,false);
out << ans << "\n";
out.close();
return 0;
}
```