#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 123;
const int inf = 1e9;
const int mod = 1e9 + 7;
const int bl = 300;
int n;
long long a[N][N];
long long b[N], c[N], mx[2], x[2], y[2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  mx[0] = mx[1] = -1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      cin >> a[i][j];
      b[i + j] += a[i][j];
      c[i - j + n] += a[i][j];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (mx[(i + j) % 2] < b[i + j] + c[i - j + n] - a[i][j]) {
        x[(i + j) % 2] = i;
        y[(i + j) % 2] = j;
        mx[(i + j) % 2] = b[i + j] + c[i - j + n] - a[i][j];
      }
    }
  }
  cout << mx[0] + mx[1] << endl;
  cout << x[0] << ' ' << y[0] << ' ' << x[1] << ' ' << y[1];
  return 0;
}

### Prompt

Please create a solution in cpp 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 int N = 2e3 + 123;
const int inf = 1e9;
const int mod = 1e9 + 7;
const int bl = 300;
int n;
long long a[N][N];
long long b[N], c[N], mx[2], x[2], y[2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  mx[0] = mx[1] = -1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      cin >> a[i][j];
      b[i + j] += a[i][j];
      c[i - j + n] += a[i][j];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (mx[(i + j) % 2] < b[i + j] + c[i - j + n] - a[i][j]) {
        x[(i + j) % 2] = i;
        y[(i + j) % 2] = j;
        mx[(i + j) % 2] = b[i + j] + c[i - j + n] - a[i][j];
      }
    }
  }
  cout << mx[0] + mx[1] << endl;
  cout << x[0] << ' ' << y[0] << ' ' << x[1] << ' ' << y[1];
  return 0;
}

```