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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Application {

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.valueOf(reader.readLine());

        int[][] matrix = new int[n][];

        for (int i = 0; i < n; i++)
            matrix[i] = parse(reader.readLine());

        long[] lDown = new long[2 * n - 1];
        long[] rDown = new long[2 * n - 1];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                lDown[i + j] += matrix[i][j];
                rDown[i - j + n - 1] += matrix[i][j];
            }
        }

        long max = 0, maxI = 0, maxJ = 0;

        for (int i = 0; i < lDown.length; i++) {
            if (lDown[i] > max) {
                max = lDown[i];
                maxI = i;
            }
        }

        max = 0;
        for (int i = 0; i < rDown.length; i++) {
            if (rDown[i] > max) {
                max = rDown[i];
                maxJ = i;
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                long val = lDown[i + j] + rDown[i - j + n - 1] - matrix[i][j];
                if (val > max) {
                    max = val;
                    maxI = i;
                    maxJ = j;
                }
            }
        }

        boolean xor = (maxI % 2 == 0) ^ (maxJ % 2 == 0);

        long totalMax = 0, totalMaxI = 0, totalMaxJ = 1;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                boolean currXor = (i % 2 == 0) ^ (j % 2 == 0);
                if (xor == currXor)
                    continue;

                long val = lDown[i + j] + rDown[i - j + n - 1] - matrix[i][j];
                if (val > totalMax) {
                    totalMax = val;
                    totalMaxI = i;
                    totalMaxJ = j;
                }
            }
        }

        totalMax += max;

        System.out.println(totalMax);
        System.out.println((maxI + 1) + " " + (maxJ + 1) + " " + (totalMaxI + 1) + " " + (totalMaxJ + 1));
    }

    private static int[] parse(String line) {
        String[] arr = line.split(" ");
        int[] res = new int[arr.length];

        for (int i = 0; i < res.length; i++)
            res[i] = Integer.valueOf(arr[i]);

        return res;
    }
}
