#include <bits/stdc++.h>
using namespace std;
int tbl[2001][2001];
long long dm[4001];
long long dt[4001];
struct trip {
  int x, y;
  long long val;
  trip() {}
  trip(int X, int Y, long long V) : x(X), y(Y), val(V) {}
};
int mabs(int x) { return x < 0 ? -x : x; }
int main() {
  int n, i, j, d;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) {
      scanf("%d", &tbl[i][j]);
    }
  }
  for (d = 1; d < 2 * n; ++d) {
    i = d > n ? n : d;
    j = d > n ? d - n + 1 : 1;
    dm[d] = 0;
    while (i && j <= n) {
      dm[d] += tbl[i][j];
      i--;
      j++;
    }
    i = d > n ? 1 : n - d + 1;
    j = d > n ? d - n + 1 : 1;
    dt[d] = 0;
    while (i <= n && j <= n) {
      dt[d] += tbl[i][j];
      i++;
      j++;
    }
  }
  long long best = -1;
  trip bestRec;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) {
      long long val = dm[i + j - 1] + dt[n - i + j] - tbl[i][j];
      if (best < val) {
        best = val;
        bestRec = trip(i, j, val);
      }
    }
  long long secondBest = -1;
  trip secondBestRec;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) {
      if (!(i == bestRec.x && j == bestRec.y) &&
          ((i + j) & 1) != ((bestRec.x + bestRec.y) & 1)) {
        long long val = dm[i + j - 1] + dt[n - i + j] - tbl[i][j];
        if (secondBest < val) {
          secondBest = val;
          secondBestRec = trip(i, j, val);
        }
      }
    }
  cout << best + secondBest << endl;
  cout << bestRec.x << " " << bestRec.y << " " << secondBestRec.x << " "
       << secondBestRec.y << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 tbl[2001][2001];
long long dm[4001];
long long dt[4001];
struct trip {
  int x, y;
  long long val;
  trip() {}
  trip(int X, int Y, long long V) : x(X), y(Y), val(V) {}
};
int mabs(int x) { return x < 0 ? -x : x; }
int main() {
  int n, i, j, d;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) {
      scanf("%d", &tbl[i][j]);
    }
  }
  for (d = 1; d < 2 * n; ++d) {
    i = d > n ? n : d;
    j = d > n ? d - n + 1 : 1;
    dm[d] = 0;
    while (i && j <= n) {
      dm[d] += tbl[i][j];
      i--;
      j++;
    }
    i = d > n ? 1 : n - d + 1;
    j = d > n ? d - n + 1 : 1;
    dt[d] = 0;
    while (i <= n && j <= n) {
      dt[d] += tbl[i][j];
      i++;
      j++;
    }
  }
  long long best = -1;
  trip bestRec;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) {
      long long val = dm[i + j - 1] + dt[n - i + j] - tbl[i][j];
      if (best < val) {
        best = val;
        bestRec = trip(i, j, val);
      }
    }
  long long secondBest = -1;
  trip secondBestRec;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) {
      if (!(i == bestRec.x && j == bestRec.y) &&
          ((i + j) & 1) != ((bestRec.x + bestRec.y) & 1)) {
        long long val = dm[i + j - 1] + dt[n - i + j] - tbl[i][j];
        if (secondBest < val) {
          secondBest = val;
          secondBestRec = trip(i, j, val);
        }
      }
    }
  cout << best + secondBest << endl;
  cout << bestRec.x << " " << bestRec.y << " " << secondBestRec.x << " "
       << secondBestRec.y << endl;
  return 0;
}

```