#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<vector<int>> A(n, vector<int>(n, 0));
  long long d1[4002] = {0};
  long long d2[4002] = {0};
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      cin >> A[i][j], d1[i + j] += A[i][j], d2[n - 1 - i + j] += A[i][j];
  pair<int, int> B1 = {0, 0}, B2 = {0, 1};
  long long b1 = 0, b2 = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if ((i + j) & 1 && d1[i + j] + d2[n - i - 1 + j] - A[i][j] > b2)
        b2 = d1[i + j] + d2[n - i - 1 + j] - A[i][j], B2 = {i, j};
      else if ((i + j) % 2 == 0 && d1[i + j] + d2[n - i - 1 + j] - A[i][j] > b1)
        b1 = d1[i + j] + d2[n - i - 1 + j] - A[i][j], B1 = {i, j};
  cout << b1 + b2 << "\n"
       << B1.first + 1 << " " << B1.second + 1 << " " << B2.first + 1 << " "
       << B2.second + 1 << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

Gargari is jealous that his friend Caisa won the game from the previous problem. He wants to prove that he is a genius.

He has a n × n chessboard. Each cell of the chessboard has a number written on it. Gargari wants to place two bishops on the chessboard in such a way that there is no cell that is attacked by both of them. Consider a cell with number x written on it, if this cell is attacked by one of the bishops Gargari will get x dollars for it. Tell Gargari, how to place bishops on the chessboard to get maximum amount of money.

We assume a cell is attacked by a bishop, if the cell is located on the same diagonal with the bishop (the cell, where the bishop is, also considered attacked by it).

Input

The first line contains a single integer n (2 ≤ n ≤ 2000). Each of the next n lines contains n integers aij (0 ≤ aij ≤ 109) — description of the chessboard.

Output

On the first line print the maximal number of dollars Gargari will get. On the next line print four integers: x1, y1, x2, y2 (1 ≤ x1, y1, x2, y2 ≤ n), where xi is the number of the row where the i-th bishop should be placed, yi is the number of the column where the i-th bishop should be placed. Consider rows are numbered from 1 to n from top to bottom, and columns are numbered from 1 to n from left to right.

If there are several optimal solutions, you can print any of them.

Examples

Input

4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1


Output

12
2 2 3 2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<vector<int>> A(n, vector<int>(n, 0));
  long long d1[4002] = {0};
  long long d2[4002] = {0};
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      cin >> A[i][j], d1[i + j] += A[i][j], d2[n - 1 - i + j] += A[i][j];
  pair<int, int> B1 = {0, 0}, B2 = {0, 1};
  long long b1 = 0, b2 = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if ((i + j) & 1 && d1[i + j] + d2[n - i - 1 + j] - A[i][j] > b2)
        b2 = d1[i + j] + d2[n - i - 1 + j] - A[i][j], B2 = {i, j};
      else if ((i + j) % 2 == 0 && d1[i + j] + d2[n - i - 1 + j] - A[i][j] > b1)
        b1 = d1[i + j] + d2[n - i - 1 + j] - A[i][j], B1 = {i, j};
  cout << b1 + b2 << "\n"
       << B1.first + 1 << " " << B1.second + 1 << " " << B2.first + 1 << " "
       << B2.second + 1 << "\n";
  return 0;
}

```