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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> a(n, vector<int>(n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cin >> a[i][j];
    }
  }
  map<pair<int, int>, long long> diag1, diag2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      int mn1 = min(i, j), mn2 = min(i, n - j - 1);
      diag1[{i - mn1, j - mn1}] += a[i][j];
      diag2[{i - mn2, j + mn2}] += a[i][j];
    }
  }
  vector<long long> sum(2);
  vector<pair<int, int>> bishops(2);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      int pos = (i + j) % 2;
      int mn1 = min(i, j), mn2 = min(i, n - j - 1);
      long long s =
          diag1[{i - mn1, j - mn1}] + diag2[{i - mn2, j + mn2}] - a[i][j];
      if (sum[pos] <= s) {
        sum[pos] = s;
        bishops[pos] = {i + 1, j + 1};
      }
    }
  }
  cout << sum[0] + sum[1] << '\n'
       << bishops[0].first << ' ' << bishops[0].second << ' '
       << bishops[1].first << ' ' << bishops[1].second << '\n';
  return 0;
}
