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
import java.io.*;
import java.util.*;

public class Main {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(br.readLine());
    int[][] aa = new int[n][n];
    long[] ipj = new long[n + n];
    long[] imj = new long[n + n];
    for (int i = 0; i < n; i++) {
      StringTokenizer st = new StringTokenizer(br.readLine());
      for (int j = 0; j < n; j++) {
        int a = Integer.parseInt(st.nextToken());
        aa[i][j] = a;
        ipj[i + j] += a;
        imj[i - j + n] += a;
      }
    }
    long max0 = -1, max1 = -1;
    int i0 = -1, j0 = -1, i1 = -1, j1 = -1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        long sum = ipj[i + j] + imj[i - j + n] - aa[i][j];
        if ((i + j) % 2 == 0 && max0 < sum) {
          max0 = sum;
          i0 = i;
          j0 = j;
        } else if ((i + j) % 2 == 1 && max1 < sum) {
          max1 = sum;
          i1 = i;
          j1 = j;
        }
      }
    System.out.println(max0 + max1);
    System.out.println((i0 + 1) + " " + (j0 + 1) + " " + (i1 + 1) + " " + (j1 + 1));
  }
}