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;
const double eps(1e-8);
int n;
long long f[2010][2010], g[2010][2010], a[2010][2010], b[2010][2010],
    x[2010][2010];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%d", &x[i][j]);
      f[i][j] = g[i][j] = a[i][j] = b[i][j] = x[i][j];
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i && j) {
        f[i][j] += f[i - 1][j - 1];
      }
  for (int i = n - 1; i >= 0; i--)
    for (int j = n - 1; j >= 0; j--)
      if (i != n - 1 && j != n - 1) {
        g[i][j] += g[i + 1][j + 1];
      }
  for (int i = 0; i < n; i++)
    for (int j = n - 1; j >= 0; j--)
      if (i && j != n - 1) {
        a[i][j] += a[i - 1][j + 1];
      }
  for (int i = n - 1; i >= 0; i--)
    for (int j = 0; j < n; j++)
      if (i != n - 1 && j) {
        b[i][j] += b[i + 1][j - 1];
      }
  long long Max1 = -1, Max2 = -1, ax, ay, bx, by;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      long long tmp = f[i][j] + g[i][j] + a[i][j] + b[i][j] - x[i][j] * 3;
      if ((i + j) & 1) {
        if (tmp > Max1) {
          Max1 = tmp;
          ax = i;
          ay = j;
        }
      } else {
        if (tmp > Max2) {
          Max2 = tmp;
          bx = i;
          by = j;
        }
      }
    }
  cout << Max1 + Max2 << endl;
  cout << ax + 1 << " " << ay + 1 << " " << bx + 1 << " " << by + 1 << endl;
  return 0;
}
