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

public class SolutionB extends Thread {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                                            InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        solve();
        out.close();
    }

    static class Cell {
        int r;
        int c;

        public Cell(int r, int c) {
            this.r = r;
            this.c = c;
        }
    }

    private static void solve() {
        int n = scanner.nextInt();
        int[][] a = new int[n][n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = scanner.nextInt();
            }
        }

        long[] diagonal1Cost = new long[2*n];

        for (int i = -n; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i + j >= 0 && i + j < n) {
                    diagonal1Cost[i+n] += a[i+j][j];
                }
            }
        }

        long[] diagonal2Cost = new long[2*n];
        for (int r = 0; r < 2 * n; r++) {
            for (int c = 0; c < n; c++) {
                if (r - c >= 0 && r - c < n) {
                    diagonal2Cost[r] += a[r-c][c];
                }
            }
        }

        long[][] c = new long[n][n];
        for (int i = 0; i < n; i++) {
            String s = "";
            for (int j = 0; j < n; j++) {
                c[i][j] = diagonal1Cost[i - j + n] + diagonal2Cost[i + j] - a[i][j];
            }
        }

        long maxEarningsWhite = -1;
        Cell bestS = null;
        
        long maxEarningsBlack = -1;
        Cell bestT = null;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if ((i + j) % 2 == 0) {
                    if (c[i][j] > maxEarningsWhite) {
                        maxEarningsWhite = c[i][j];
                        bestS = new Cell(i, j);
                    } 
                } else {
                    if (c[i][j] > maxEarningsBlack) {
                        maxEarningsBlack = c[i][j];
                        bestT = new Cell(i, j);
                    }
                }
            }
        }
        
        out.println(maxEarningsWhite + maxEarningsBlack);
        out.println(bestS.r+1 + " " + (bestS.c+1) + " " + (bestT.r+1) + " " + (bestT.c+1));
    }

    //REMINDERS:
    //- CHECK FOR INTEGER-OVERFLOW BEFORE SUBMITTING

    //- CAN U BRUTEFORCE OVER SOMETHING, TO MAKE IT EASIER TO CALCULATE THE SOLUTION
}