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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class T463C {

	public void solve(int n, int[][] a) {
		
		long[][] dp1 = new long[n][n];
		long[][] dp2 = new long[n][n];
		for (int i = 0; i < n; i ++) {
			for (int j = 0; j < n; j ++) {
				if (i - 1 >= 0 && j - 1 >= 0)
					dp1[i][j] = dp1[i - 1][j - 1] + a[i][j];
				else
					dp1[i][j] = a[i][j];
				
				if (i - 1 >= 0 && j + 1 < n)
					dp2[i][j] = dp2[i - 1][j + 1] + a[i][j];
				else
					dp2[i][j] = a[i][j];
			}
		}
		
		int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
		long max1 = 0, max2 = 0, ans = 0;

		for (int i = 0; i < n; i ++) {
			for (int j = 0; j < n; j ++) {
				if (i >= j)
					ans = dp1[n - 1][j + n - i - 1];
				else
					ans = dp1[i + n - j - 1][n - 1];
				
				if (i + j >= n - 1)
					ans += dp2[n - 1][i + j - n + 1];
				else
					ans += dp2[j + i][0];
				
				ans -= a[i][j];
				
				if (max1 <= ans && (i + j) % 2 == 0) {
					max1 = ans;
					x1 = i;
					y1 = j;
				}
				if (max2 <= ans && (i + j) % 2 == 1) {
					max2 = ans;
					x2 = i;
					y2 = j;
				}
			}
		}

		System.out.println(max1 + max2);
		System.out.println((x1 + 1) + " " + (y1 + 1) + " " + (x2 + 1) + " " + (y2 + 1));
	}
	
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(nextToken());
		}

		long nextLong() {
			return Long.parseLong(nextToken());
		}

		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}
	
	public static void main(String[] args) {
		T463C t = new T463C();
		
		FastScanner in = new FastScanner();
		int n = in.nextInt();
		int[][] a = new int[n][n];
		for (int i = 0; i < n; i ++) {
			for (int j = 0; j < n; j ++)
				a[i][j] = in.nextInt();
		}
		t.solve(n, a);
		
	}

}
