#include <bits/stdc++.h>
using namespace std;
int main() {
  long long m1 = 0, m2 = 0, p = 1, q = 1, r = 1, s = 2, n, g;
  scanf("%I64d", &n);
  vector<long long> a(2 * n - 1), b = a;
  vector<vector<long long>> v(n, vector<long long>(n));
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) {
      scanf("%I64d", &v[i][j]);
      a[j - i + n - 1] += v[i][j];
      b[i + j] += v[i][j];
    }
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) {
      if (a[j - i + n - 1] + b[i + j] - v[i][j] > m1 && (i + j + 1) % 2)
        m1 = a[j - i + n - 1] + b[i + j] - v[i][j], p = i + 1, q = j + 1;
      if (a[j - i + n - 1] + b[i + j] - v[i][j] > m2 && (i + j) % 2)
        m2 = a[j - i + n - 1] + b[i + j] - v[i][j], r = i + 1, s = j + 1;
    }
  g = m1 + m2;
  printf("%I64d \n", g);
  printf("%I64d %I64d %I64d %I64d", p, q, r, s);
}

### 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;
int main() {
  long long m1 = 0, m2 = 0, p = 1, q = 1, r = 1, s = 2, n, g;
  scanf("%I64d", &n);
  vector<long long> a(2 * n - 1), b = a;
  vector<vector<long long>> v(n, vector<long long>(n));
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) {
      scanf("%I64d", &v[i][j]);
      a[j - i + n - 1] += v[i][j];
      b[i + j] += v[i][j];
    }
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) {
      if (a[j - i + n - 1] + b[i + j] - v[i][j] > m1 && (i + j + 1) % 2)
        m1 = a[j - i + n - 1] + b[i + j] - v[i][j], p = i + 1, q = j + 1;
      if (a[j - i + n - 1] + b[i + j] - v[i][j] > m2 && (i + j) % 2)
        m2 = a[j - i + n - 1] + b[i + j] - v[i][j], r = i + 1, s = j + 1;
    }
  g = m1 + m2;
  printf("%I64d \n", g);
  printf("%I64d %I64d %I64d %I64d", p, q, r, s);
}

```