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.InputStreamReader;

public class SummingDiagonals {
    static int n;
    static long arr[][];

    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(bf.readLine());
        arr = new long[n][n];
        String[] l;
        for (int i = 0; i < arr.length; i++) {
            l = bf.readLine().split(" ");
            for (int j = 0; j < arr.length; j++) {
                arr[i][j] = Integer.parseInt(l[j]);
            }
        }
        long[][] primaryDigSums = new long[n][n];
        long[][] secondaryDigSums = new long[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                primaryDigSums[i][j] = arr[i][j];
                secondaryDigSums[i][j] = arr[i][(n - 1) - j];
            }
        }
        long maxOdd = -1;
        int x1 = 0;
        int y1 = 0;
        long maxEven = -1;
        int x2 = 0;
        int y2 = 1;

        sumPrimaryDiagonals(primaryDigSums);
        sumPrimaryDiagonals(secondaryDigSums);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                arr[i][j] = primaryDigSums[i][j]
                        + secondaryDigSums[i][(n - 1) - j] - arr[i][j];
                if ((i + j) % 2 == 0) {
                    if (arr[i][j] > maxEven) {
                        maxEven = arr[i][j];
                        x1 = i + 1;
                        y1 = j + 1;
                    }
                } else {
                    if (arr[i][j] > maxOdd) {
                        maxOdd = arr[i][j];
                        x2 = i + 1;
                        y2 = j + 1;
                    }
                }
            }
        }
        System.out.println(maxEven + maxOdd);
        System.out.println(x1 + " " + y1 + " " + x2 + " " + y2);
    }

    /**
     * Sums all the primary diagonals (result in res)
     * 
     * @param arr
     * @param res
     */
    static void sumPrimaryDiagonals(long[][] res) {
        for (int j = 0; j < n; j++) {
            int y = 1;
            int x = j + 1;
            for (; x < n; x++, y++) {
                res[y][x] += res[y - 1][x - 1];
                if (j != 0)
                    res[x][y] += res[x - 1][y - 1];
            }
            long sum = res[y - 1][x - 1];
            long sum2 = res[x - 1][y - 1];
            y = y - 2;
            for (x = x - 2; x >= 0 && y >= 0; x--, y--) {
                res[y][x] = sum;
                if (j != 0)
                    res[x][y] = sum2;
            }

        }
    }

}
