#include <bits/stdc++.h>
using namespace std;
const long long D = 2007;
long long A[D][D], TL[D][D], BL[D][D], TR[D][D], BR[D][D];
void PrintMat(long long M[D][D], long long n) {
  for (long long i = 1; i <= n; i++, cout << endl)
    for (long long j = 1; j <= n; j++) cout << M[i][j] << " ";
  cout << "-------\n";
}
void Solve() {
  long long n;
  cin >> n;
  memset(A, 0, sizeof(A));
  memset(TL, 0, sizeof(TL));
  memset(BL, 0, sizeof(BL));
  memset(TR, 0, sizeof(TR));
  memset(BR, 0, sizeof(BR));
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++) cin >> A[i][j];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      TL[i][j] = TR[i][j] = A[i][j];
      if (i - 1 >= 1 && j - 1 >= 1) TL[i][j] += TL[i - 1][j - 1];
      if (i - 1 >= 1 && j + 1 <= n) TR[i][j] += TR[i - 1][j + 1];
    }
  }
  for (long long i = n; i >= 1; i--) {
    for (long long j = 1; j <= n; j++) {
      BL[i][j] = BR[i][j] = A[i][j];
      if (i + 1 <= n && j - 1 >= 1) BL[i][j] += BL[i + 1][j - 1];
      if (i + 1 <= n && j + 1 <= n) BR[i][j] += BR[i + 1][j + 1];
    }
  }
  long long mx1 = 0, mx2 = 0, x1, y1, x2, y2, f = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++, f = !f) {
      long long tmp = TL[i][j] + TR[i][j] + BL[i][j] + BR[i][j] - 3 * A[i][j];
      if (f == 0 && tmp >= mx1) {
        mx1 = tmp, x1 = i, y1 = j;
      }
      if (f == 1 && tmp >= mx2) {
        mx2 = tmp, x2 = i, y2 = j;
      }
    }
    if (n % 2 == 0) f = !f;
  }
  cout << mx1 + mx2 << '\n'
       << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  Solve();
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const long long D = 2007;
long long A[D][D], TL[D][D], BL[D][D], TR[D][D], BR[D][D];
void PrintMat(long long M[D][D], long long n) {
  for (long long i = 1; i <= n; i++, cout << endl)
    for (long long j = 1; j <= n; j++) cout << M[i][j] << " ";
  cout << "-------\n";
}
void Solve() {
  long long n;
  cin >> n;
  memset(A, 0, sizeof(A));
  memset(TL, 0, sizeof(TL));
  memset(BL, 0, sizeof(BL));
  memset(TR, 0, sizeof(TR));
  memset(BR, 0, sizeof(BR));
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++) cin >> A[i][j];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      TL[i][j] = TR[i][j] = A[i][j];
      if (i - 1 >= 1 && j - 1 >= 1) TL[i][j] += TL[i - 1][j - 1];
      if (i - 1 >= 1 && j + 1 <= n) TR[i][j] += TR[i - 1][j + 1];
    }
  }
  for (long long i = n; i >= 1; i--) {
    for (long long j = 1; j <= n; j++) {
      BL[i][j] = BR[i][j] = A[i][j];
      if (i + 1 <= n && j - 1 >= 1) BL[i][j] += BL[i + 1][j - 1];
      if (i + 1 <= n && j + 1 <= n) BR[i][j] += BR[i + 1][j + 1];
    }
  }
  long long mx1 = 0, mx2 = 0, x1, y1, x2, y2, f = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++, f = !f) {
      long long tmp = TL[i][j] + TR[i][j] + BL[i][j] + BR[i][j] - 3 * A[i][j];
      if (f == 0 && tmp >= mx1) {
        mx1 = tmp, x1 = i, y1 = j;
      }
      if (f == 1 && tmp >= mx2) {
        mx2 = tmp, x2 = i, y2 = j;
      }
    }
    if (n % 2 == 0) f = !f;
  }
  cout << mx1 + mx2 << '\n'
       << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  Solve();
  return 0;
}

```