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;

public class GAGRI_AND_BISHOP {

	public static void main(String[] args) throws NumberFormatException, IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		int n = Integer.parseInt(br.readLine());

		long[][] matrix = new long[n][n];
		long[] forward = new long[2 * n];
		long[] backward = new long[2 * n];

		String[] elements;
		for (int i = 1; i <= n; i++) {
			elements = br.readLine().split(" ");
			for (int j = 1; j <= n; j++) {
				matrix[i - 1][j - 1] = Long.parseLong(elements[j - 1]);
			}
		}

		int offset = n;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				forward[i + j - 1] += matrix[i - 1][j - 1];
				backward[j - i - 1 + offset] += matrix[i - 1][j - 1];
			}
		}

		long maxI = -1;
		int maxIR = 1, maxIC = 1;
		long maxJ = -1;
		int maxJR = 1, maxJC = 2;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				long sum = forward[i + j - 1] + backward[j - i - 1 + offset] - matrix[i - 1][j - 1];

				if ((i + j) % 2 == 0) {
					if (sum > maxI) {
						maxI = sum;

						maxIR = i;
						maxIC = j;
					}
				} else {
					if (sum > maxJ) {
						maxJ = sum;

						maxJR = i;
						maxJC = j;
					}
				}
			}
		}

		System.out.println(maxI + maxJ);
		System.out.println(maxIR + " " + maxIC + " " + maxJR + " " + maxJC);

	}

}
