#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long n, tmp, mx1 = 0, mx2 = 0;
  cin >> n;
  vector<long long> diag1(2 * n - 1), diag2(2 * n - 1), intrs[2 * n - 1];
  vector<pair<int, int> > ptint[2 * n - 1];
  pair<int, int> pt1, pt2;
  for (int i = 0; i < 2 * n - 1; ++i)
    intrs[i].resize(2 * n - 1), ptint[i].resize(2 * n - 1),
        fill(ptint[i].begin(), ptint[i].end(), make_pair(-1, -1));
  fill(diag1.begin(), diag1.end(), 0);
  fill(diag2.begin(), diag2.end(), 0);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      cin >> tmp;
      diag1[i + j] += tmp;
      diag2[n - 1 - j + i] += tmp;
      intrs[i + j][n - 1 - j + i] = tmp;
      ptint[i + j][n - 1 - j + i] = make_pair(i + 1, j + 1);
    }
  int i1 = -1, j1 = -1;
  for (int i = 0; i < 2 * n - 1; ++i)
    for (int j = 0; j < 2 * n - 1; ++j)
      if (diag1[i] + diag2[j] - intrs[i][j] >= mx1 && ptint[i][j].first != -1) {
        pt1 = ptint[i][j];
        mx1 = diag1[i] + diag2[j] - intrs[i][j];
        i1 = i;
        j1 = j;
      }
  for (int i = 0; i < 2 * n - 1; ++i)
    for (int j = 0; j < 2 * n - 1; ++j)
      if (diag1[i] + diag2[j] - intrs[i][j] >= mx2 && ptint[i][j].first != -1 &&
          i != i1 && j != j1 && ptint[i][j1].first == -1 &&
          ptint[i1][j].first == -1) {
        pt2 = ptint[i][j];
        mx2 = diag1[i] + diag2[j] - intrs[i][j];
      }
  cout << mx1 + mx2 << '\n';
  cout << pt1.first << ' ' << pt1.second << ' ' << pt2.first << ' '
       << pt2.second;
}

### Prompt

In cpp, your task is to solve 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_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long n, tmp, mx1 = 0, mx2 = 0;
  cin >> n;
  vector<long long> diag1(2 * n - 1), diag2(2 * n - 1), intrs[2 * n - 1];
  vector<pair<int, int> > ptint[2 * n - 1];
  pair<int, int> pt1, pt2;
  for (int i = 0; i < 2 * n - 1; ++i)
    intrs[i].resize(2 * n - 1), ptint[i].resize(2 * n - 1),
        fill(ptint[i].begin(), ptint[i].end(), make_pair(-1, -1));
  fill(diag1.begin(), diag1.end(), 0);
  fill(diag2.begin(), diag2.end(), 0);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      cin >> tmp;
      diag1[i + j] += tmp;
      diag2[n - 1 - j + i] += tmp;
      intrs[i + j][n - 1 - j + i] = tmp;
      ptint[i + j][n - 1 - j + i] = make_pair(i + 1, j + 1);
    }
  int i1 = -1, j1 = -1;
  for (int i = 0; i < 2 * n - 1; ++i)
    for (int j = 0; j < 2 * n - 1; ++j)
      if (diag1[i] + diag2[j] - intrs[i][j] >= mx1 && ptint[i][j].first != -1) {
        pt1 = ptint[i][j];
        mx1 = diag1[i] + diag2[j] - intrs[i][j];
        i1 = i;
        j1 = j;
      }
  for (int i = 0; i < 2 * n - 1; ++i)
    for (int j = 0; j < 2 * n - 1; ++j)
      if (diag1[i] + diag2[j] - intrs[i][j] >= mx2 && ptint[i][j].first != -1 &&
          i != i1 && j != j1 && ptint[i][j1].first == -1 &&
          ptint[i1][j].first == -1) {
        pt2 = ptint[i][j];
        mx2 = diag1[i] + diag2[j] - intrs[i][j];
      }
  cout << mx1 + mx2 << '\n';
  cout << pt1.first << ' ' << pt1.second << ' ' << pt2.first << ' '
       << pt2.second;
}

```