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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		long arr[][] = new long[n][n];
		long dig1[] = new long[n+n];
		long dig2[] = new long[n+n];
		for(int i=0;i<n;i++){
			StringTokenizer st = new StringTokenizer(br.readLine());
			for(int j=0;j<n;j++){
				int a = Integer.parseInt(st.nextToken());
				arr[i][j]=a;
				dig1[i+j] += a;
				dig2[i-j+n] += a;
			}
		}
		long max1=-1;
		long max2=-1;
		int i1=0,j1=0,i2=0,j2=0;
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				long sum = dig1[i+j] + dig2[i-j+n] - arr[i][j];
				if((i+j)%2 == 0 && sum>max1){
					max1 = sum;
					i1 = i;
					j1 = j;
				}
				else
				     if((i+j)%2 == 1 && sum>max2){
					max2 = sum;
					i2 = i;
					j2 = j;
				}
			}
		}
		
		System.out.println(max1+max2);
		System.out.println((i1+1)+" "+ (j1+1) + " "+ (i2+1) + " " + (j2+1));

	}


}