#include <bits/stdc++.h>
using namespace std;
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const pair<Arg1, Arg2>& a) {
  return out << "(" << a.first << " , " << a.second << ")";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const vector<Arg1>& a) {
  out << "[";
  for (const auto& i : a) out << i << " , ";
  return out << "]";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const set<Arg1>& a) {
  out << "{";
  for (const auto& i : a) out << i << " , ";
  return out << "}";
}
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const map<Arg1, Arg2>& a) {
  out << "<";
  for (const auto& i : a) out << i << " , ";
  return out << ">";
}
template <typename T, size_t N>
typename enable_if<!is_same<typename remove_cv<T>::type, char>::value,
                   ostream&>::type
operator<<(ostream& out, T (&a)[N]) {
  out << '[';
  for (size_t i = 0; i < N; ++i) out << a[i] << " , ";
  out << ']';
  return out;
}
long long int a[2005][2005], d1[2 * 2005] = {0}, d2[2 * 2005] = {0},
                                    ans[2] = {0}, first[2], second[2];
void upd(long long int i, long long int u, long long int v, long long int sm) {
  if (sm >= ans[i]) {
    ans[i] = sm;
    first[i] = u;
    second[i] = v;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j;
  cin >> n;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[n + i - j] += a[i][j];
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      upd((i + j) & 1, i, j, d1[i + j] + d2[n + i - j] - a[i][j]);
  cout << ans[0] + ans[1] << "\n";
  cout << first[0] + 1 << " " << second[0] + 1 << " " << first[1] + 1 << " "
       << second[1] + 1;
  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;
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const pair<Arg1, Arg2>& a) {
  return out << "(" << a.first << " , " << a.second << ")";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const vector<Arg1>& a) {
  out << "[";
  for (const auto& i : a) out << i << " , ";
  return out << "]";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const set<Arg1>& a) {
  out << "{";
  for (const auto& i : a) out << i << " , ";
  return out << "}";
}
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const map<Arg1, Arg2>& a) {
  out << "<";
  for (const auto& i : a) out << i << " , ";
  return out << ">";
}
template <typename T, size_t N>
typename enable_if<!is_same<typename remove_cv<T>::type, char>::value,
                   ostream&>::type
operator<<(ostream& out, T (&a)[N]) {
  out << '[';
  for (size_t i = 0; i < N; ++i) out << a[i] << " , ";
  out << ']';
  return out;
}
long long int a[2005][2005], d1[2 * 2005] = {0}, d2[2 * 2005] = {0},
                                    ans[2] = {0}, first[2], second[2];
void upd(long long int i, long long int u, long long int v, long long int sm) {
  if (sm >= ans[i]) {
    ans[i] = sm;
    first[i] = u;
    second[i] = v;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j;
  cin >> n;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[n + i - j] += a[i][j];
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      upd((i + j) & 1, i, j, d1[i + j] + d2[n + i - j] - a[i][j]);
  cout << ans[0] + ans[1] << "\n";
  cout << first[0] + 1 << " " << second[0] + 1 << " " << first[1] + 1 << " "
       << second[1] + 1;
  return 0;
}

```