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 sum[4010];
long long int dif[4010];
long long int ch[2010][2010];
int main() {
  int n;
  cin >> n;
  n++;
  for (int i = (1); i < (n); i++) {
    for (int j = (1); j < (n); j++) {
      scanf("%d", &ch[i][j]);
      sum[i + j] += ch[i][j];
      dif[i - j + n - 1] += ch[i][j];
    }
  }
  long long int sol[2] = {-1, -1};
  int x[2] = {-1, -1};
  int y[2] = {-1, -1};
  for (int i = (1); i < (n); i++) {
    for (int j = (1); j < (n); j++) {
      long long int cand = sum[i + j] + dif[i - j + n - 1] - ch[i][j];
      if (cand > sol[(i + j) & 1]) {
        sol[(i + j) & 1] = cand;
        x[(i + j) & 1] = i;
        y[(i + j) & 1] = j;
      }
    }
  }
  cout << (sol[0] + sol[1]) << endl;
  cout << x[0] << " " << y[0] << " " << x[1] << " " << y[1] << endl;
  return 0;
}
