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 int MAX = 2005;
int n;
int t[MAX][MAX];
long long diag1[2 * MAX], diag2[2 * MAX];
int main() {
  scanf(" %d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf(" %d", &t[i][j]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i + j < n) diag1[i] += (long long)t[j][i + j];
      if (i - j >= 0) diag2[i] += (long long)t[j][i - j];
      if (i != 0 && i + j < n) {
        diag1[n + i - 1] += (long long)t[i + j][j];
        diag2[n + i - 1] += (long long)t[i + j][n - j - 1];
      }
    }
  }
  tuple<long long, int, int> bishop1, bishop2;
  get<1>(bishop1) = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      long long aux = -t[i][j];
      if (i > j) {
        aux += diag1[n - 1 + (i - j)];
      } else {
        aux += diag1[j - i];
      }
      aux += diag2[i + j];
      if ((i + j) % 2 == 0 && aux > get<0>(bishop1))
        bishop1 = make_tuple(aux, i, j);
      if ((i + j) % 2 == 1 && aux > get<0>(bishop2))
        bishop2 = make_tuple(aux, i, j);
    }
  }
  long long b1, b2;
  int x1, x2, y1, y2;
  tie(b1, x1, y1) = bishop1;
  tie(b2, x2, y2) = bishop2;
  printf("%lld\n%d %d %d %d\n", b1 + b2, x1 + 1, y1 + 1, x2 + 1, y2 + 1);
}
