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 a[2003][2003];
unordered_map<int, long long> pr;
long long se[4004];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int i, j;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      cin >> a[i][j];
      pr[j - i] += a[i][j];
      se[i + j] += a[i][j];
    }
  }
  long long oddp = 0;
  long long evenp = 0;
  pair<long long, long long> pairOdd = {1, 1};
  pair<long long, long long> pairEven = {1, 2};
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      long long psum = pr[j - i];
      long long ssum = se[i + j];
      long long tans = psum + ssum - a[i][j];
      if ((i + j) % 2 == 0) {
        if (oddp < tans) {
          oddp = tans;
          pairEven = {i, j};
        }
      } else {
        if (evenp < tans) {
          evenp = tans;
          pairOdd = {i, j};
        }
      }
    }
  }
  cout << oddp + evenp << "\n";
  cout << pairEven.first << " " << pairEven.second << " " << pairOdd.first
       << " " << pairOdd.second << "\n";
  return 0;
}
