#include <bits/stdc++.h>
using namespace std;
int n;
long long sum1[4005], sum2[4005];
int a[2005][2005];
long long f(int i, int j) { return sum1[i - j + n] + sum2[i + j] - a[i][j]; }
int main() {
  while (scanf("%d", &n) == 1) {
    memset(sum1, 0, sizeof(sum1));
    memset(sum2, 0, sizeof(sum2));
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        int x;
        scanf("%d", &x);
        a[i][j] = x;
        sum1[i - j + n] += x;
        sum2[i + j] += x;
      }
    int ansx[10], ansy[10];
    ansx[1] = 1;
    ansy[1] = 2;
    ansx[2] = 1;
    ansy[2] = 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        if ((i + j) % 2 == 1 && f(i, j) > f(ansx[1], ansy[1])) {
          ansx[1] = i;
          ansy[1] = j;
        }
        if ((i + j) % 2 == 0 && f(i, j) > f(ansx[2], ansy[2])) {
          ansx[2] = i;
          ansy[2] = j;
        }
      }
    cout << f(ansx[1], ansy[1]) + f(ansx[2], ansy[2]) << endl;
    cout << ansx[1] << ' ' << ansy[1] << endl;
    cout << ansx[2] << ' ' << ansy[2] << endl;
  }
  return 0;
}

### 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>
using namespace std;
int n;
long long sum1[4005], sum2[4005];
int a[2005][2005];
long long f(int i, int j) { return sum1[i - j + n] + sum2[i + j] - a[i][j]; }
int main() {
  while (scanf("%d", &n) == 1) {
    memset(sum1, 0, sizeof(sum1));
    memset(sum2, 0, sizeof(sum2));
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        int x;
        scanf("%d", &x);
        a[i][j] = x;
        sum1[i - j + n] += x;
        sum2[i + j] += x;
      }
    int ansx[10], ansy[10];
    ansx[1] = 1;
    ansy[1] = 2;
    ansx[2] = 1;
    ansy[2] = 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        if ((i + j) % 2 == 1 && f(i, j) > f(ansx[1], ansy[1])) {
          ansx[1] = i;
          ansy[1] = j;
        }
        if ((i + j) % 2 == 0 && f(i, j) > f(ansx[2], ansy[2])) {
          ansx[2] = i;
          ansy[2] = j;
        }
      }
    cout << f(ansx[1], ansy[1]) + f(ansx[2], ansy[2]) << endl;
    cout << ansx[1] << ' ' << ansy[1] << endl;
    cout << ansx[2] << ' ' << ansy[2] << endl;
  }
  return 0;
}

```