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 Abc {
    static int arr[][];
    static long sum[][];
    static int n;
    public static void main(String[] args) {
        FastReader sc = new FastReader();
        n=sc.nextInt();
        arr=new int[n][n];sum=new long[n][n];
        long sec[]=new long[2*n],prim[]=new long[2*n];

        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                arr[i][j]=sc.nextInt();
                sec[i+j]+=arr[i][j];
                prim[i-j+n]+=arr[i][j];
            }
        }
        long maxE=-1,maxO=-1;
        int ans[]=new int[4];
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                long sum=prim[i-j+n]+sec[i+j]-arr[i][j];
                if ((i+j)%2==0){
                    if (maxE<sum){
                        maxE=sum;
                        ans[0]=i;ans[1]=j;
                    }
                }else {
                    if (maxO<sum){
                        maxO=sum;
                        ans[2]=i;ans[3]=j;
                    }
                }
            }
        }
        System.out.println(maxE+maxO);
        for (int i=0;i<4;i++)ans[i]++;
        System.out.println(ans[0]+" "+ans[1]+" "+ans[2]+" "+ans[3]);
    }


    static 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;
        }
    }
}