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.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static int mod=(int)1e9+7;
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        int n = sc.nextInt();
        int[][] arr = new int[n + 1][n + 1];
        long[] d1=new long[2*n+1],d2=new long[2*n+1],res=new long[2];
        int x[] = new int[2], y[] = new int[2];

        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                arr[i][j]=sc.nextInt();
                d1[i+j]+=arr[i][j];
                d2[i-j+n]+=arr[i][j];
            }
        }
        res[0]=res[1]=-1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                int pos=(i+j)&1;
                long sum=d1[i+j]+d2[i-j+n]-arr[i][j];
                if(sum>res[pos]){
                    res[pos]=sum;
                    x[pos]=i;
                    y[pos]=j;
                }
            }
        }
        System.out.println(res[0]+res[1]);
        System.out.println(x[0]+" "+y[0]+" "+x[1]+" "+y[1]);
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;

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

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

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

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
