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;
int n, m;
long long a[2010][2010];
long long trai[4 * 2010], phai[4 * 2010];
struct arr {
  long long val;
  int x, y;
};
void solve() {
  int i, j, xchan, ychan, xle, yle;
  arr temp;
  long long maxchan, maxle;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      trai[i + j - 1] += a[i][j];
      phai[i + (n - j + 1) - 1] += a[i][j];
    }
  maxchan = -1;
  maxle = -1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      temp.val = trai[i + j - 1] + phai[i + (n - j + 1) - 1] - a[i][j];
      temp.x = i;
      temp.y = j;
      if ((i + j) % 2 == 0) {
        if (temp.val > maxchan) {
          maxchan = temp.val;
          xchan = temp.x;
          ychan = temp.y;
        }
      } else {
        if (temp.val > maxle) {
          maxle = temp.val;
          xle = temp.x;
          yle = temp.y;
        }
      }
    }
  cout << maxchan + maxle << endl;
  cout << xchan << " " << ychan << " " << xle << " " << yle << endl;
}
void input() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) scanf("%I64d", &a[i][j]);
  solve();
}
int main() {
  input();
  fclose(stdin);
  fclose(stdout);
  return 0;
}
