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.io.PrintWriter;
import java.util.HashMap;

public class GargariAndBishops {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int n = Integer.parseInt(br.readLine());
		int[][] grid = new int[n][n];
		for(int i = 0 ; i < n ; i++){
			String[] in = br.readLine().split("\\s+");
			for(int j = 0; j < n; j++) {
				grid[i][j] = Integer.parseInt(in[j]);
			}
		}
		int id = 0;
		HashMap<Integer, Long> diag1 = new HashMap<>(), diag2 = new HashMap<>();
		for( int k = 0 ; k < n ; k++, id++){
			long sum = 0;
			for( int j = 0 ; j <= k ; j++){
				int i = k - j;
				sum+=grid[i][j];
			}
			diag1.put(id, sum);
		}
		for( int k = n - 2 ; k >= 0 ; k--, id++){
			long sum = 0;
			for( int j = 0 ; j <= k ; j++){
				int i = k - j;
				sum+=grid[n-j-1][n-i-1];
			}
			diag1.put(id, sum);
		}
		
		for(int i = n-1 ; i >= 0 ; i--){
			long sum = 0;
			int temp = 0;
			for(int j = i ; j < n ; j++, temp++){
				sum+=grid[j][temp];
			}
			diag2.put(i, sum);
		}
		for(int i = 1 ; i < n ; i++){
			long sum = 0;
			int temp = 0;
			for(int j = i ; j < n ; j++, temp++){
				sum+=grid[temp][j];
			}
			diag2.put(-i, sum);
		}
		long max1 = 0, max2 = 0;
		int r1 = 1, c1 = 1, r2 = 1, c2 = 2;
		for(int i = 0 ; i < n ; i++){
			for(int j = 0 ; j < n ; j++){
				long curr = diag1.get(i+j) + diag2.get(i-j) - grid[i][j];
				if((i+j)%2 == 1){
					if(curr > max1){
						max1 = curr;
						r2 = i+1; c2 = j+1;
					}
				}
				if((i+j)%2 == 0){
					if(curr > max2){
						max2 = curr;
						r1 = i+1; c1 = j+1;
					}
				}
			}
		}
		out.printf("%d\n%d %d %d %d",max1+max2, r1, c1, r2, c2);
		out.close();
	}
}


