#include <bits/stdc++.h>
const int SIZE = 4e3 + 5;
using namespace std;
int arr[SIZE][SIZE];
long long Right[SIZE];
long long Left[SIZE];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> arr[i][j];
      Right[i - j + n] += arr[i][j];
      Left[i + j] += arr[i][j];
    }
  }
  long long Ans1 = -1;
  int a, b;
  long long Ans2 = -1;
  int c, d;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      long long Sum = Right[i - j + n] + Left[i + j] - arr[i][j];
      if (Sum > Ans1 && (i + j) % 2 == 1) {
        Ans1 = Sum;
        a = i + 1;
        b = j + 1;
      }
      if (Sum > Ans2 && (i + j) % 2 == 0) {
        Ans2 = Sum;
        c = i + 1;
        d = j + 1;
      }
    }
  }
  cout << Ans1 + Ans2 << '\n';
  printf("%d %d %d %d", a, b, c, d);
}

### Prompt

Your challenge is to write a cpp solution to 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>
const int SIZE = 4e3 + 5;
using namespace std;
int arr[SIZE][SIZE];
long long Right[SIZE];
long long Left[SIZE];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> arr[i][j];
      Right[i - j + n] += arr[i][j];
      Left[i + j] += arr[i][j];
    }
  }
  long long Ans1 = -1;
  int a, b;
  long long Ans2 = -1;
  int c, d;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      long long Sum = Right[i - j + n] + Left[i + j] - arr[i][j];
      if (Sum > Ans1 && (i + j) % 2 == 1) {
        Ans1 = Sum;
        a = i + 1;
        b = j + 1;
      }
      if (Sum > Ans2 && (i + j) % 2 == 0) {
        Ans2 = Sum;
        c = i + 1;
        d = j + 1;
      }
    }
  }
  cout << Ans1 + Ans2 << '\n';
  printf("%d %d %d %d", a, b, c, d);
}

```