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 INF = 1000000001;
const double EPS = 10e-9;
int n, pl[2345][2345];
long long diagonals[4567][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= (n); ++i)
    for (int j = 1; j <= (n); ++j) cin >> pl[i][j];
  for (int i = 1; i <= (n); ++i)
    for (int j = 1; j <= (n); ++j) {
      diagonals[i + j][0] += pl[i][j];
      diagonals[i - j + n][1] += pl[i][j];
    }
  int x1, y1, x2, y2;
  long long max1 = -1;
  long long max2 = -1;
  long long pomoc;
  for (int i = 1; i <= (n); ++i)
    for (int j = 1; j <= (n); ++j) {
      if ((i + j) & 1) {
        pomoc = diagonals[i + j][0] + diagonals[i - j + n][1];
        pomoc -= pl[i][j];
        if (pomoc > max1) {
          max1 = pomoc;
          x1 = i;
          y1 = j;
        }
      } else {
        pomoc = diagonals[i + j][0] + diagonals[i - j + n][1];
        pomoc -= pl[i][j];
        if (pomoc > max2) {
          max2 = pomoc;
          x2 = i;
          y2 = j;
        }
      }
    }
  cout << max1 + max2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
  return 0;
}
