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 Bishops {
    public static void main(String[]args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int [][] board = new int[n][n];
        for (int i = 0; i < n; i++) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            for (int j = 0; j < n; j++) {
                board[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        long [][] diagsRight = new long[n][n];
        long [][] diagsLeft = new long[n][n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i - 1 >= 0 && j+1 < n) {
                    diagsLeft[i][j] += diagsLeft[i-1][j+1] + board[i][j];
                } else {
                    diagsLeft[i][j] = board[i][j];
                }

                if (i - 1 >= 0 && j-1 >= 0) {
                    diagsRight[i][j] += diagsRight[i-1][j-1] + board[i][j];
                } else {
                    diagsRight[i][j] = board[i][j];
                }
            }
        }
        
        long max1 = 0;
        int x1 = 1;
        int y1 = 1;
        int x2 = 1;
        int y2 = 2;
        long max2 = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int d1 = n-1-i;
                int d2 = n-1-j;
                int minD = Math.min(d1, d2);
                long sum1Right = diagsRight[i+minD][j+minD];
                d1 = n-1-i;
                d2 = j;
                minD = Math.min(d1, d2);
                long sum1Left = diagsLeft[i+minD][j-minD];
                long total = sum1Left+sum1Right-board[i][j];
                if ((i+j)%2 == 0) {
                    if (total > max1) {
                        max1 = total;
                        x1 = i+1;
                        y1 = j+1;
                    }
                } else {
                    if (total > max2) {
                        max2 = total;
                        x2 = i+1;
                        y2 = j+1;
                    }
                }
            }
        }
        System.out.println(max1+max2);
        System.out.printf("%d %d %d %d\n", x1, y1, x2, y2);
    }
}
