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>
namespace patch {
template <typename T>
std::string to_string(const T& n) {
  std::ostringstream stm;
  stm << n;
  return stm.str();
}
}  // namespace patch
using namespace std;
long long int RD[5000], LD[5000], arr[2000][2000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      cin >> arr[i][j];
      RD[i - j + 2500] += arr[i][j];
      LD[i + j] += arr[i][j];
    }
  }
  long long int ans[2] = {}, X1, X2, Y1, Y2;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (ans[(i + j) % 2] <= RD[i - j + 2500] + LD[i + j] - arr[i][j]) {
        ans[(i + j) % 2] = RD[i - j + 2500] + LD[i + j] - arr[i][j];
        (i + j) % 2 == 0 ? (X1 = i, Y1 = j) : (X2 = i, Y2 = j);
      }
    }
  }
  cout << ans[0] + ans[1] << '\n';
  cout << X1 + 1 << " " << Y1 + 1 << " " << X2 + 1 << " " << Y2 + 1;
  return 0;
}
