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.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class C {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(bf.readLine());
        int[][] a = new int[n][n];
        for(int i=0; i<n; i++) {
            StringTokenizer st = new StringTokenizer(bf.readLine());
            for(int j=0; j<n; j++) a[i][j] = Integer.parseInt(st.nextToken());
        }
        long[] sum = new long[2*n];
        long[] difference = new long[2*n];
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                sum[i+j] += a[i][j];
                difference[n+i-j] += a[i][j];
            }
        }
       long max1 = -1; int i1 = -1; int j1 = -1;
        long max2 = -1; int i2 = -1; int j2 = -1;
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                if((i+j)%2 == 0) {
                    long val = sum[i+j] + difference[n+i-j] - a[i][j];
                    if(val > max1) {
                        max1 = val;
                        i1 = i+1;
                        j1 = j+1;
                    }
                }
                else {
                    long val = sum[i+j] + difference[n+i-j] - a[i][j];
                    if(val > max2) {
                        max2 = val;
                        i2 = i+1;
                        j2 = j+1;
                    }
                }
            }
        }
        System.out.println((max1 + max2));
        System.out.println(i1 + " " + j1 + " " + i2 + " " + j2);
        // out.close(); System.exit(0);
    }
}