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

/**
 * Created by chinh on 11/17/14.
 */
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.StringTokenizer;

public class R264D2C {

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int[][] a = new int[n][n];
        HashMap<Integer, Long> sum1 = new HashMap<Integer, Long>();
        HashMap<Integer, Long> sum2 = new HashMap<Integer, Long>();
        for(int i=0; i<n; i++){
            StringTokenizer st = new StringTokenizer(br.readLine());
            for(int j=0; j<n; j++){
                a[i][j] = Integer.parseInt(st.nextToken());
                sum1.put(i-j, sum1.containsKey(i-j) ? sum1.get(i-j)+a[i][j] : a[i][j]);
                sum2.put(i+j, sum2.containsKey(i+j) ? sum2.get(i+j)+a[i][j] : a[i][j]);
            }
        }
        long maxOdd=-1, maxEven=-1;
        int ie=0, je=0, io=0, jo=0;
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                long cur = sum1.get(i-j)+sum2.get(i+j)-a[i][j];
                if((i+j)%2==0){
                    if(maxEven<cur){
                        maxEven = cur;
                        ie = i+1;
                        je = j+1;
                    }
                } else{
                    if(maxOdd<cur){
                        maxOdd = cur;
                        io = i+1;
                        jo = j+1;
                    }
                }
            }
        }
        System.out.println(maxOdd+maxEven);
        System.out.println(ie+" "+je+" "+io+" "+jo);
    }
}