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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Locale;
import java.util.StringTokenizer;

public class C {
	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException {
		Locale.setDefault(Locale.US);
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(
				System.out)));
		int n = nextInt();
		long[][] a = new long[n + 2][n + 2];
		long a1[] = new long[4005], a2[] = new long[4005];
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++) {
				a[i][j] = nextInt();
				a1[i + j] += a[i][j];
				a2[i - j + n] += a[i][j];
			}
		long ans1 = 0, ans2 = 0;
		int x1 = 1, y1 = 1, x2 = 1, y2 = 2;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				long s = a1[i + j] + a2[i - j + n] - a[i][j];
				if ((i + j) % 2 == 0) {
					if (s > ans1) {
						ans1 = s;
						x1 = i;
						y1 = j;
					}
				} else {
					if (s > ans2) {
						ans2 = s;
						x2 = i;
						y2 = j;
					}
				}
			}
		}
		System.out.println(ans1 + ans2);
		System.out.println(x1 + " " + y1 + " " + x2 + " " + y2);
	}

	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	private static String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}
