#include <bits/stdc++.h>
using namespace std;
long long n, grid[2500][2500], R[10005], C[10005];
vector<pair<long long, pair<long long, long long> > > A, B;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++)
      cin >> grid[i][j], R[i + j] += grid[i][j], C[i - j + 5000] += grid[i][j];
  long long Min = (long long)1e18;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      long long Z = R[i + j] + C[i - j + 5000] - grid[i][j];
      Min = min(Min, Z);
      if (B.size() < 5000 || Z >= Min)
        B.push_back(make_pair(R[i + j] + C[i - j + 5000] - grid[i][j],
                              make_pair(i, j)));
    }
  }
  sort(B.begin(), B.end());
  reverse(B.begin(), B.end());
  long long a, b, x, y;
  long long Max = 0;
  for (long long i = 0; i < min((long long)B.size(), 5000LL); i++) {
    for (long long j = 0; j < min((long long)B.size(), 3000LL); j++) {
      if (B[i].first + B[j].first >= Max &&
          ((B[i].second.first + B[i].second.second - B[j].second.first +
            B[j].second.second)) %
              2 &&
          ((B[i].second.first - B[i].second.second -
            (B[j].second.first - B[j].second.second))) %
              2) {
        a = B[i].second.first, b = B[i].second.second;
        x = B[j].second.first, y = B[j].second.second;
        Max = B[i].first + B[j].first;
      }
    }
  }
  cout << Max << endl;
  cout << a + 1 << ' ' << b + 1 << ' ' << x + 1 << ' ' << y + 1 << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
long long n, grid[2500][2500], R[10005], C[10005];
vector<pair<long long, pair<long long, long long> > > A, B;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++)
      cin >> grid[i][j], R[i + j] += grid[i][j], C[i - j + 5000] += grid[i][j];
  long long Min = (long long)1e18;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      long long Z = R[i + j] + C[i - j + 5000] - grid[i][j];
      Min = min(Min, Z);
      if (B.size() < 5000 || Z >= Min)
        B.push_back(make_pair(R[i + j] + C[i - j + 5000] - grid[i][j],
                              make_pair(i, j)));
    }
  }
  sort(B.begin(), B.end());
  reverse(B.begin(), B.end());
  long long a, b, x, y;
  long long Max = 0;
  for (long long i = 0; i < min((long long)B.size(), 5000LL); i++) {
    for (long long j = 0; j < min((long long)B.size(), 3000LL); j++) {
      if (B[i].first + B[j].first >= Max &&
          ((B[i].second.first + B[i].second.second - B[j].second.first +
            B[j].second.second)) %
              2 &&
          ((B[i].second.first - B[i].second.second -
            (B[j].second.first - B[j].second.second))) %
              2) {
        a = B[i].second.first, b = B[i].second.second;
        x = B[j].second.first, y = B[j].second.second;
        Max = B[i].first + B[j].first;
      }
    }
  }
  cout << Max << endl;
  cout << a + 1 << ' ' << b + 1 << ' ' << x + 1 << ' ' << y + 1 << endl;
  return 0;
}

```