#include <bits/stdc++.h>
using namespace std;
long long int sol[2];
pair<long long int, long long int> v[2];
long long int a[2000 + 1][2000 + 1], sum1[2000 * 3], sum2[2000 * 3];
void change(long long int x, long long int i, long long int j,
            long long int val) {
  if (val > sol[x]) {
    v[x].first = i;
    v[x].second = j;
    sol[x] = val;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      cin >> a[i][j];
      sum1[i + j] += a[i][j];
      sum2[i - j + n] += a[i][j];
    }
  }
  sol[0] = sol[1] = -1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      change((i + j) & 1, i, j, sum1[i + j] + sum2[i - j + n] - a[i][j]);
    }
  }
  cout << sol[0] + sol[1] << endl;
  cout << v[0].first << " " << v[0].second << endl;
  cout << v[1].first << " " << v[1].second << endl;
  return 0;
}

### 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;
long long int sol[2];
pair<long long int, long long int> v[2];
long long int a[2000 + 1][2000 + 1], sum1[2000 * 3], sum2[2000 * 3];
void change(long long int x, long long int i, long long int j,
            long long int val) {
  if (val > sol[x]) {
    v[x].first = i;
    v[x].second = j;
    sol[x] = val;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      cin >> a[i][j];
      sum1[i + j] += a[i][j];
      sum2[i - j + n] += a[i][j];
    }
  }
  sol[0] = sol[1] = -1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      change((i + j) & 1, i, j, sum1[i + j] + sum2[i - j + n] - a[i][j]);
    }
  }
  cout << sol[0] + sol[1] << endl;
  cout << v[0].first << " " << v[0].second << endl;
  cout << v[1].first << " " << v[1].second << endl;
  return 0;
}

```