#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-4;
const long double PI = acos((long double)-1.0);
const long long int MOD = 1e9 + 7;
long long int powmod(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int debug = 1;
template <typename T, typename S>
void DBG(pair<T, S> p) {
  if (debug) cout << p.first << " " << p.second << endl;
}
template <typename T>
void DBG(vector<T> v) {
  if (debug) {
    for (T t : v) cout << t << "  ";
    cout << endl;
  }
}
template <typename T>
void DBG(T t) {
  if (debug) cout << t << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  vector<long long int> diag(2 * n - 1, 0), anti(2 * n - 1, 0);
  vector<vector<int>> v(n, vector<int>(n));
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      cin >> v[i][j];
      diag[j - i + (n - 1)] += v[i][j];
      anti[i + j] += v[i][j];
    }
  }
  vector<long long int> best(2, 0);
  vector<pair<int, int>> p(2);
  multimap<long long int, pair<int, int>> mm, mm2;
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      long long int value = (diag[j - i + (n - 1)] + anti[i + j] - v[i][j]);
      if (value >= best[(i + j) % 2]) {
        best[(i + j) % 2] = value;
        p[(i + j) % 2] = make_pair(i + 1, j + 1);
      }
    }
  }
  DBG(accumulate(best.begin(), best.end(), 0LL));
  DBG(vector<int>{p[0].first, p[0].second, p[1].first, p[1].second});
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 double EPS = 1e-4;
const long double PI = acos((long double)-1.0);
const long long int MOD = 1e9 + 7;
long long int powmod(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int debug = 1;
template <typename T, typename S>
void DBG(pair<T, S> p) {
  if (debug) cout << p.first << " " << p.second << endl;
}
template <typename T>
void DBG(vector<T> v) {
  if (debug) {
    for (T t : v) cout << t << "  ";
    cout << endl;
  }
}
template <typename T>
void DBG(T t) {
  if (debug) cout << t << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  vector<long long int> diag(2 * n - 1, 0), anti(2 * n - 1, 0);
  vector<vector<int>> v(n, vector<int>(n));
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      cin >> v[i][j];
      diag[j - i + (n - 1)] += v[i][j];
      anti[i + j] += v[i][j];
    }
  }
  vector<long long int> best(2, 0);
  vector<pair<int, int>> p(2);
  multimap<long long int, pair<int, int>> mm, mm2;
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      long long int value = (diag[j - i + (n - 1)] + anti[i + j] - v[i][j]);
      if (value >= best[(i + j) % 2]) {
        best[(i + j) % 2] = value;
        p[(i + j) % 2] = make_pair(i + 1, j + 1);
      }
    }
  }
  DBG(accumulate(best.begin(), best.end(), 0LL));
  DBG(vector<int>{p[0].first, p[0].second, p[1].first, p[1].second});
}

```