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 Main {

	public static void main(String[] args) throws IOException {
		BufferedReader bi = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(bi.readLine());
		long[][] a = new long[n][n];
		for(int i=0;i<n;i++) {
			String s = bi.readLine();
			String[] sa = s.split("\\s");
			for(int j=0;j<n;j++) {
				a[i][j] = Integer.parseInt(sa[j]);
			}
		}
		int m = 2 * n - 1;
		long[] sum1 = new long[m];// /
		long[] sum2 = new long[m];// \
		for(int i=0;i<n;i++) {
			for(int j=0;j<=i;j++) {
				sum1[i] += a[j][i-j];
			}
		}
		for(int i=1;i<n;i++) {
			for(int j=0;i+j<n;j++) {
				sum1[n+i-1] += a[i+j][n-1-j];
			}
		}
		for(int i=0;i<n;i++) {
			for(int j=0;n-1-i+j<n;j++) {
				sum2[i] += a[n-1-i+j][j];
			}
		}
		for(int i=1;i<n;i++) {
			for(int j=0;i+j<n;j++) {
				sum2[n+i-1] += a[j][i+j];
			}
		}
		//System.out.println(Arrays.toString(sum1));
		//System.out.println(Arrays.toString(sum2));
		long max1 = -1;
		long max2 = -1;
		int i1 = 0,j1 = 0,i2 = 0,j2 = 0;
		for(int i=0;i<n;i++) {
			for(int j=0;j<n;j++) {
				long num = sum1[i+j]+sum2[n-1+j-i]-a[i][j];
				//System.out.println(i + "," + j + "," + num);
				if ((i+j)%2==0) {
					if (num > max1) {
						i1 = i;
						j1 = j;
						max1 = num;
					}
				}else{
					if (num > max2) {
						i2 = i;
						j2 = j;
						max2 = num;
					}
				}
			}
		}
		System.out.println(max1 + max2);
		System.out.println((i1+1) + " " + (j1+1) + " " + (i2+1) + " " + (j2+1));
	}

}
