#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
const long long MAXN = 2e3 + 69;
const long long MOD = 1e9 + 7;
const long long BASE = 11;
int n, arr[MAXN][MAXN];
pair<long long, pair<int, int>> res[2], ans[MAXN * MAXN];
map<int, long long> diag[2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      cin >> arr[i][j], diag[0][i - j] += arr[i][j],
          diag[1][i + j] += arr[i][j];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      ans[i * n + j].first = diag[0][i - j] + diag[1][i + j] - arr[i][j],
                  ans[i * n + j].second.first = i + 1,
                  ans[i * n + j].second.second = j + 1;
  for (int i = 0; i < n * n; i++)
    if (ans[i].first >=
        res[(ans[i].second.first + ans[i].second.second) % 2].first) {
      res[(ans[i].second.first + ans[i].second.second) % 2].first =
          ans[i].first;
      res[(ans[i].second.first + ans[i].second.second) % 2].second.first =
          ans[i].second.first;
      res[(ans[i].second.first + ans[i].second.second) % 2].second.second =
          ans[i].second.second;
    }
  cout << res[0].first + res[1].first << "\n";
  for (int i = 0; i < 2; i++)
    cout << res[i].second.first << " " << res[i].second.second << " ";
  return cout << '\n', 0;
}

### Prompt

Generate a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
const long long MAXN = 2e3 + 69;
const long long MOD = 1e9 + 7;
const long long BASE = 11;
int n, arr[MAXN][MAXN];
pair<long long, pair<int, int>> res[2], ans[MAXN * MAXN];
map<int, long long> diag[2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      cin >> arr[i][j], diag[0][i - j] += arr[i][j],
          diag[1][i + j] += arr[i][j];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      ans[i * n + j].first = diag[0][i - j] + diag[1][i + j] - arr[i][j],
                  ans[i * n + j].second.first = i + 1,
                  ans[i * n + j].second.second = j + 1;
  for (int i = 0; i < n * n; i++)
    if (ans[i].first >=
        res[(ans[i].second.first + ans[i].second.second) % 2].first) {
      res[(ans[i].second.first + ans[i].second.second) % 2].first =
          ans[i].first;
      res[(ans[i].second.first + ans[i].second.second) % 2].second.first =
          ans[i].second.first;
      res[(ans[i].second.first + ans[i].second.second) % 2].second.second =
          ans[i].second.second;
    }
  cout << res[0].first + res[1].first << "\n";
  for (int i = 0; i < 2; i++)
    cout << res[i].second.first << " " << res[i].second.second << " ";
  return cout << '\n', 0;
}

```