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
#include <bits/stdc++.h>
using namespace std;
long long int n, m, dollar[2048][2048];
long long int mdoll;
long long int odd, even, ox, oy, ex, ey;
pair<long long int, long long int> val[2048][2048];
stack<pair<long long int, long long int> > stk;
void check() {
  long long int a, b;
  long long int temp;
  for (a = 1; a <= n; a++) {
    for (b = 1; b <= n; b++) {
      if ((a + b) % 2 == 1) {
        temp = val[a][b].first + val[a][b].second - dollar[a][b];
        if (temp >= odd) {
          odd = temp;
          ox = a;
          oy = b;
        }
      } else {
        temp = val[a][b].first + val[a][b].second - dollar[a][b];
        if (temp >= even) {
          even = temp;
          ex = a;
          ey = b;
        }
      }
    }
  }
  return;
}
void total_in_diagonal() {
  long long int a, x, y, sq;
  long long int sum;
  m = 2 * n - 1;
  for (a = 1; a <= m; a++) {
    sum = 0;
    if (a <= n) {
      x = 1;
      y = a;
      sq = a;
    } else {
      x = 1 + (a % n);
      y = n;
      sq = n - (a % n);
    }
    while (sq--) {
      sum += dollar[x][y];
      stk.push(make_pair(x, y));
      x++;
      y--;
    }
    while (!stk.empty()) {
      val[(stk.top()).first][(stk.top()).second].first = sum;
      stk.pop();
    }
  }
  for (a = 1; a <= m; a++) {
    sum = 0;
    if (a <= n) {
      y = 1;
      x = n - a + 1;
      sq = a;
    } else {
      y = 1 + (a % n);
      x = 1;
      sq = n - (a % n);
    }
    while (sq--) {
      sum += dollar[x][y];
      stk.push(make_pair(x, y));
      y++;
      x++;
    }
    while (!stk.empty()) {
      val[(stk.top()).first][(stk.top()).second].second = sum;
      stk.pop();
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  long long int a, b;
  cin >> n;
  for (a = 1; a <= n; a++) {
    for (b = 1; b <= n; b++) {
      cin >> dollar[a][b];
    }
  }
  total_in_diagonal();
  check();
  cout << odd + even << '\n';
  cout << ox << " " << oy << " " << ex << " " << ey << '\n';
  return 0;
}
