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 a_ij (0 ≤ a_ij ≤ 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 x_i is the number of the row where the i-th bishop should be placed, y_i 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.

Example

Input

4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1

Output

12
2 2 3 2
此题思路就是计算每一个位置所能收获的价值，用一个变量来存取最大值。利用的重点每一条主对角线的x-y是一个定值而每一个副对角线之和x+y是一个定值；

#include<stdio.h>
const int N = 2005;
typedef long long LL;
LL Map[N][N];
LL L[N*2];//主对角线
LL R[N*2];//副对角线
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            scanf("%I64d",&Map[i][j]);
            L[i-j+n] += Map[i][j];//主对角线下x，y之和为定值n为确保不会出现负值
            R[i+j] += Map[i][j];//副对角线x，y之差为定值
        }
    }
//显然比下文常规算法更简短易懂
/*for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                m=0;
                if(i>=j)
                {
                    for(int x=0;x<n-i+j;x++)
                    {
                        m=m+a[i-j+x][x];
                    }
                }
                else
                {
                    for(int x=0;x<n-j+i;x++)
                    {
                        m=m+a[x][j-i+x];
                    }
                }
                if(i+j<=n)
                {
                    for(int x=0;x<i+j+1;x++)
                    {
                        m=m+a[x][i+j-x];
                    }
                }
                else
                {
                    for(int x=n-1;x>i+j-n;x--)
                    {
                        m=m+a[x][i+j-x];
                    }
                }
                m=m-a[i][j];
                if(m>m1)
                {
                    m1=m;
                    x1=i+1;
                    y1=j+1;
                }
            }
        }
*/
    int x1 = 1, x2 = 1, y1 = 1, y2 = 2;
    LL max1 = 0, max2 = 0;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            LL tmp = L[i+j] + R[i-j+n] - Map[i][j];
            if((i+j) % 2 == 0 && tmp > max1) {
                max1 = tmp;
                x1 = i;
                y1 = j;
            }
            if((i + j) % 2 == 1 && tmp > max2) {
                max2 = tmp;
                x2 = i;
                y2 = j;
            }
        }
    }
    printf("%I64d\n", max1 + max2);
    printf("%d %d %d %d\n", x1, y1, x2, y2);
}