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.StringTokenizer;

public class pr463C {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        int n = in.nextInt();
        int[][] grid = new int[n][n];

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

        PrintWriter out = new PrintWriter(System.out);
        solve(n, grid, out);
        out.flush();
        out.close();
    }

    private static void solve(int n, int[][] grid, PrintWriter out) {
        long[] d1 = new long[2*n];
        long[] d2 = new long[2*n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                d1[i + j] += grid[i][j];
                d2[i - j + n] += grid[i][j];
            }
        }

        int[] x = {1, 1};
        int[] y = {1, 2};
        long[] sol = new long[2];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                long val = d1[i + j] + d2[i - j + n] - grid[i][j];
                update(i, j, val, x, y, sol, (i+j)&1);
            }
        }

        out.println(sol[0] + sol[1]);
        for (int i = 0; i < 2; i++) {
            out.print(x[i] + " " + y[i] + " ");
        }
    }
    private static void update(int i, int j, long val, int[] x, int[] y, long[] sol, int c) {
        if(val > sol[c]){
            sol[c] = val;
            x[c] = i+1;
            y[c]  = j+1;
        }
    }

    static class FastScanner {
        StringTokenizer st;
        BufferedReader br;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        boolean hasNext() throws IOException {
            return br.ready() || (st != null && st.hasMoreTokens());
        }

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

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

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

        String nextLine() throws IOException {
            return br.readLine();
        }

        boolean hasNextLine() throws IOException {
            return br.ready();
        }

    }
}
