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.nio.Buffer;
import java.util.Arrays;
import java.util.Scanner;

public class ProblemC {
	static int[][] grid;
	static int n;

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(bf.readLine());
		long[] dp = new long[2 * n];
		long[] dp1 = new long[2*n];
		Arrays.fill(dp, 0);
		Arrays.fill(dp1, 0);
		grid = new int[n][n];
		for (int i = 0; i < n; i++){
			String [] q = bf.readLine().trim().split("\\s+");
			for (int j = 0; j < n; j++) {
				int w = Integer.parseInt(q[j]);
				dp[i + j] += w;
				dp1[n-i+j] += w;
				grid[i][j] = w;
			}
		}
		long max1 = Long.MIN_VALUE;
		int loci = 0;
		int locj = 0;
		long max2 = Long.MIN_VALUE;
		int loci2 = 0;
		int locj2 = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) {
				long maxa = dp[i + j];
				maxa += dp1[n-i+j];
				maxa -= grid[i][j];
				if ((i + j) % 2 == 0) {
					if (maxa > max1) {
						max1 = maxa;
						loci = i;
						locj = j;
					}
				} else {
					if (maxa > max2) {
						max2 = maxa;
						loci2 = i;
						locj2 = j;
					}
				}
			}
		System.out.println(max1 + max2);
		System.out.println((loci + 1) + " " + (locj + 1) + " " + (loci2 + 1)
				+ " " + (locj2 + 1));
	}
}