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
#include <bits/stdc++.h>
using namespace std;
const int N = 2 * (1e3) + 10;
long long int tab[N][N];
long long int preP[N][N], preS[N][N];
int main() {
  int n;
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> tab[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      preP[i][j] = tab[i][j];
      preS[i][j] = tab[i][j];
      if (i > 0 && j > 0) {
        preP[i][j] += preP[i - 1][j - 1];
      }
      if (i > 0 && j < n - 1) {
        preS[i][j] += preS[i - 1][j + 1];
      }
    }
  }
  int posi1, posj1, posi2, posj2;
  long long int max1 = -1, max2 = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int m = min(n - i, n - j) - 1;
      int k = min(n - i - 1, j);
      long long int c = preP[i + m][j + m] + preS[i + k][j - k] - tab[i][j];
      if ((i + j) % 2 == 0) {
        if (c > max1) {
          posi1 = i + 1;
          posj1 = j + 1;
          max1 = c;
        }
      } else {
        if (c > max2) {
          posi2 = i + 1;
          posj2 = j + 1;
          max2 = c;
        }
      }
    }
  }
  cout << max1 + max2 << '\n';
  cout << posi1 << ' ' << posj1 << ' ' << posi2 << ' ' << posj2 << '\n';
  return 0;
}
