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 max_n = 2000;
long long int n;
long long int mas[max_n][max_n] = {}, d[2][max_n * 2 - 1] = {};
int main() {
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      scanf("%d", &mas[i][j]);
      d[0][i + j] += mas[i][j];
      d[1][i + (n - 1) - j] += mas[i][j];
    }
  int maxpi[2], maxpj[2];
  long long int maxs[2] = {-1, -1};
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      long long int sum = d[0][i + j] + d[1][i + (n - 1) - j] - mas[i][j];
      int odd = (i + j) % 2;
      if (sum > maxs[odd]) {
        maxs[odd] = sum;
        maxpi[odd] = i + 1;
        maxpj[odd] = j + 1;
      }
    }
  cout << maxs[0] + maxs[1] << endl;
  cout << maxpi[0] << ' ' << maxpj[0] << ' ';
  cout << maxpi[1] << ' ' << maxpj[1];
}
