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.*;
import java.util.*;

public class R264_Div2_C {
	static BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st = new StringTokenizer("");

	public static void main(String[] args) throws Exception {
		int n = readInt();
      
      int[][] board = new int[n][n];
      for(int i = 0; i < n; ++i)
         for(int j = 0; j < n; ++j)
            board[i][j] = readInt();
      
      long[] se = new long[2*n-1];
      long[] sw = new long[2*n-1];
      for(int i = 1-n; i < n; ++i) {
         int x = Math.max(0, i);
         int y = Math.max(0, -i);
         while(x < n && y < n)
            se[i+n-1] += board[y++][x++];
         x = Math.max(0, -i);
         y = Math.min(n-1, n-1-i);
         while(x < n && y >= 0)
            sw[i+n-1] += board[y--][x++];
      }
      
      long[] best = {0, 0};
      int[][] pos = new int[2][2];
      for(int i = 1-n; i < n; ++i) {
         int x = Math.max(0, i);
         int y = Math.max(0, -i);
         for(int j = 0; j < n-Math.abs(i); ++j) {
            long temp = se[i+n-1]+sw[2*n-2-Math.abs(i)-2*j]-board[y++][x++];
            if(temp >= best[(x+y)%2]) {
               best[(x+y)%2] = temp;
               pos[(x+y)%2][0] = y;
               pos[(x+y)%2][1] = x;
            }
         }
      }
      
      System.out.println(best[0] + best[1]);
      System.out.println(pos[0][0]+" "+pos[0][1]+" "+pos[1][0]+" "+pos[1][1]);
	}

	static String readString() throws Exception {
		while(!st.hasMoreTokens())
			st = new StringTokenizer(stdin.readLine());
		return st.nextToken();
	}
	static int readInt() throws Exception {
		return Integer.parseInt(readString());
	}
}
