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
#include <bits/stdc++.h>
using namespace std;
long long a[3000][3000];
long long fl[3000][3000];
long long fr[3000][3000];
int main() {
  int n;
  scanf("%d", &n);
  int i, j;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) scanf("%I64d", &a[i][j]);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) fl[i][j] = fl[i - 1][j - 1] + a[i][j];
  for (i = n - 1; i >= 1; i--)
    for (j = n - 1; j >= 1; j--) fl[i][j] = fl[i + 1][j + 1];
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) fr[i][j] = fr[i - 1][j + 1] + a[i][j];
  for (i = n - 1; i >= 1; i--)
    for (j = n; j >= 2; j--) fr[i][j] = fr[i + 1][j - 1];
  long long maxx = 0, maxi, maxj, minn = 0, mini, minj;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        if (maxx <= fl[i][j] + fr[i][j] - a[i][j]) {
          maxx = fl[i][j] + fr[i][j] - a[i][j];
          maxi = i;
          maxj = j;
        }
      } else {
        if (minn <= fl[i][j] + fr[i][j] - a[i][j]) {
          minn = fl[i][j] + fr[i][j] - a[i][j];
          mini = i;
          minj = j;
        }
      }
    }
  }
  printf("%I64d\n", maxx + minn);
  printf("%I64d %I64d %I64d %I64d\n", maxi, maxj, mini, minj);
  return 0;
}
