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;
long long int mat[2005][2005], principal[2005][2005], secundaria[2005][2005];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) cin >> mat[i][j];
  for (int j = 0; j < n; j++) {
    long long int aux = 0;
    for (int k = 0; k < n && j + k < n; k++) aux += mat[0 + k][j + k];
    for (int k = 0; k < n && j + k < n; k++) principal[0 + k][j + k] = aux;
  }
  for (int i = 1; i < n; i++) {
    long long int aux = 0;
    for (int k = 0; i + k < n && k < n; k++) aux += mat[i + k][0 + k];
    for (int k = 0; i + k < n && k < n; k++) principal[i + k][0 + k] = aux;
  }
  for (int j = 0; j < n; j++) {
    long long int aux = 0;
    for (int k = 0; 0 + k < n && j - k >= 0; k++) aux += mat[0 + k][j - k];
    for (int k = 0; 0 + k < n && j - k >= 0; k++)
      secundaria[0 + k][j - k] = aux;
  }
  for (int i = 1; i < n; i++) {
    long long int aux = 0;
    for (int k = 0; i + k < n && n - 1 - k >= 0; k++)
      aux += mat[i + k][n - 1 - k];
    for (int k = 0; i + k < n && n - 1 - k >= 0; k++)
      secundaria[i + k][n - 1 - k] = aux;
  }
  long long int b1 = 0, b2 = 0;
  long long int x1 = 1, y1 = 1, x2 = 1, y2 = 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2 == 0 &&
          principal[i][j] + secundaria[i][j] - mat[i][j] > b1) {
        b1 = principal[i][j] + secundaria[i][j] - mat[i][j];
        x1 = i + 1;
        y1 = j + 1;
      }
      if ((i + j) % 2 == 1 &&
          principal[i][j] + secundaria[i][j] - mat[i][j] > b2) {
        b2 = principal[i][j] + secundaria[i][j] - mat[i][j];
        x2 = i + 1;
        y2 = j + 1;
      }
    }
  cout << b1 + b2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  return 0;
}
