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.*;


public class C {	
	BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter out;
    
	public void solve() throws IOException {				
		int N = nextInt();
		long[][] A = new long[N][N];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                A[i][j] = nextLong();
            }
        }
        long[] diag1 = new long[2 * N - 1];
        long[] diag2 = new long[2 * N - 1];

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                diag1[N-1-i+j] += A[i][j];
                diag2[i+j] += A[i][j];
            }
        }

        long[] maxSum = new long[2];
        int[][] ans = new int[2][];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                int parity = (i + j) % 2;
                long sum = diag1[N-1-i+j] + diag2[i+j] - A[i][j];
                if (sum >= maxSum[parity]) {
                    maxSum[parity] = sum;
                    ans[parity] = new int[] {i, j};
                }
            }
        }
        out.println(maxSum[0] + maxSum[1]);
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                out.print((ans[i][j]+1) + " ");
            }
        }
        out.println();
	}
	
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		new C().run();
	}
	
	public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            out = new PrintWriter(System.out);
            solve();
            reader.close();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

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

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

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

}
