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.lang.*;
import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;

        int n = Integer.parseInt(br.readLine());
        int[][] grid = new int[n][n];

        for(int i = 0 ; i < n ; i++) {
            st = new StringTokenizer(br.readLine());
            for(int j = 0 ; j < n ; j++) {
                grid[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        long[] forward = new long[2*n - 1];
        long[] backward = new long[2*n - 1];
        for(int i = 0 ; i < n ; i++) {
            for(int j = 0 ; j < n ; j++) {
                forward[i+j] += grid[i][j];
                backward[i-j + n-1] += grid[i][j];
            }
        }

        long maxA = Integer.MIN_VALUE , maxB = Integer.MIN_VALUE;
        int k = 0 , xa = -1 , ya = -1 , xb = -1 , yb = -1;

        for(int i = 0 ; i < n ; i++) {
            for(int j = i % 2 == 0 ? 0 : n-1 ; i % 2 == 0 ? j < n : j >= 0 ; j = i % 2 == 0 ? j + 1 : j - 1) {
                long pts = forward[i+j] + backward[i-j + n-1] - grid[i][j];
                if(k % 2 == 0) {
                    if(maxA < pts) {
                        maxA = pts;
                        xa = i+1;
                        ya = j+1;
                    }
                }
                else {
                    if(maxB < pts) {
                        maxB = pts;
                        xb = i+1;
                        yb = j+1;
                    }
                }
                k++;
            }
        }

        System.out.println(maxA + maxB);
        System.out.println(xa + " " + ya + " " + xb + " "  + yb);

    }

}