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.*;

public class C {

    static StreamTokenizer st;
    static int n;
    static int[][] a;

    public static void main(String[] args) throws IOException {
        st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        read();
        long[] mainDiagonal = mainDiagonalsSum();
        long[] secondaryDiagonal = secondaryDiagolanSum();
        long b1 = -1, b2 = -1;
        int r1 = 0, r2 = 0, c1 = 0, c2 = 0;
        for (int r = 0; r < n; r++) {
            for (int c = 0; c < n; c++) {
                int main = c - r + n - 1;
                int second = c + r;
                if (main % 2 == 1) {
                    if (mainDiagonal[main] + secondaryDiagonal[second] - a[r][c] > b1) {
                        b1 = mainDiagonal[main] + secondaryDiagonal[second] - a[r][c];
                        r1 = r + 1;
                        c1 = c + 1;
                    }
                } else if (mainDiagonal[main] + secondaryDiagonal[second] - a[r][c] > b2) {
                    b2 = mainDiagonal[main] + secondaryDiagonal[second] - a[r][c];
                    r2 = r + 1;
                    c2 = c + 1;
                }
            }
        }
        System.out.println(b1 + b2);
        System.out.printf("%d %d %d %d%n", r1, c1, r2, c2);
    }

    static long[] mainDiagonalsSum() {
        long[] mainDiagonal = new long[2 * n - 1];
        mainDiagonal[n - 1] = diagonalSum(0, 0);
        for (int i = 1; i < n; i++) {
            mainDiagonal[-i + n - 1] = diagonalSum(i, 0);
            mainDiagonal[i + n - 1] = diagonalSum(0, i);
        }
        return mainDiagonal;
    }

    static long diagonalSum(int r, int c) {
        long res = 0;
        for (int i = r, j = c; i < n && j < n; i++, j++) {
            res += (long) a[i][j];
        }
        return res;
    }

    static long sDiagonalSum(int r, int c) {
        long res = 0;
        for (int i = r, j = c; i < n && j >= 0; i++, j--) {
            res += (long) a[i][j];
        }
        return res;
    }

    static void read() throws IOException {
        n = ri();
        a = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = ri();
            }
        }
    }

    static int ri() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    static String rs() throws IOException {
        st.nextToken();
        return st.sval;
    }

    static long[] secondaryDiagolanSum() {
        long[] res = new long[2 * n - 1];
        res[n - 1] = sDiagonalSum(0, n - 1);
        for (int i = 1; i < n; i++) {
            res[n - 1 - i] = sDiagonalSum(0, n - 1 - i);
            res[n - 1 + i] = sDiagonalSum(i, n - 1);
        }
        return res;
    }
}
