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.*;
import java.math.*;

public class p463c {
    public static void main(String[] args)throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        long[][] a = new long[n][n];
        for(int i = 0; i < n; i++){
            String[] str = br.readLine().split(" ");
            for(int j = 0; j < n; j++){
                a[i][j] = Long.parseLong(str[j]);
            }
        }
        long[] diagonal1 = new long[2 * n - 1];
        long[] diagonal2 = new long[2 * n - 1];
        long[][] ans = new long[n][n];
        long x1 = 0, y1 = 1, x2 = 0, y2 = 0;
        long max1 = 0, max2 = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                diagonal1[i - j + n - 1] += a[i][j];
                diagonal2[i + j] += a[i][j];
            }
        }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                ans[i][j] += diagonal1[i - j + n - 1];
                ans[i][j] += diagonal2[i + j];
                ans[i][j] -= a[i][j];
            }
        }
        for(int i = 0; i < n; i ++) {
            for(int j = 0; j < n; j++) {
                if ((i + j) % 2 == 0 && ans[i][j] > max1) {
                    x2 = i;
                    y2 = j;
                    max1 = ans[i][j];
                } else if ((i + j) % 2 == 1 && ans[i][j] > max2) {
                    x1 = i;
                    y1 = j;
                    max2 = ans[i][j];
                }
            }
        }
        System.out.println(max1 + max2);
        System.out.println((x1 + 1) + " " + (y1 + 1) + " " + (x2 + 1) + " " + (y2 + 1));
    }
}