#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long N;
  cin >> N;
  vector<vector<long long>> a(N + 1, vector<long long>(N + 1));
  vector<long long> d1(N << 1), d2(N << 1);
  for (long long i = 1; i <= N; ++i)
    for (long long j = 1; j <= N; ++j) {
      cin >> a[i][j];
      d1[i + j - 1] += a[i][j];
      d2[i - j + N] += a[i][j];
    }
  vector<long long> sol(2);
  vector<pair<long long, long long>> coords(2);
  for (long long i = 1; i <= N; ++i)
    for (long long j = 1; j <= N; ++j) {
      long long index = (i + j) & 1;
      long long cand = d1[i + j - 1] + d2[i - j + N] - a[i][j];
      if (cand >= sol[index]) {
        sol[index] = cand;
        coords[index] = make_pair(i, j);
      }
    }
  cout << sol[0] + sol[1] << '\n';
  for (const pair<long long, long long> &p : coords)
    cout << p.first << ' ' << p.second << ' ';
  cout << '\n';
}

### Prompt

Your challenge is to write 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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long N;
  cin >> N;
  vector<vector<long long>> a(N + 1, vector<long long>(N + 1));
  vector<long long> d1(N << 1), d2(N << 1);
  for (long long i = 1; i <= N; ++i)
    for (long long j = 1; j <= N; ++j) {
      cin >> a[i][j];
      d1[i + j - 1] += a[i][j];
      d2[i - j + N] += a[i][j];
    }
  vector<long long> sol(2);
  vector<pair<long long, long long>> coords(2);
  for (long long i = 1; i <= N; ++i)
    for (long long j = 1; j <= N; ++j) {
      long long index = (i + j) & 1;
      long long cand = d1[i + j - 1] + d2[i - j + N] - a[i][j];
      if (cand >= sol[index]) {
        sol[index] = cand;
        coords[index] = make_pair(i, j);
      }
    }
  cout << sol[0] + sol[1] << '\n';
  for (const pair<long long, long long> &p : coords)
    cout << p.first << ' ' << p.second << ' ';
  cout << '\n';
}

```