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, a[2007][2007], m[4007], sec[4007], res[2007][2007];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) cin >> a[i][j];
  }
  pair<long long int, long long int> wmax = {0, 0}, bmax = {0, 1};
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) {
      m[j - i + 2000] += a[i][j];
      sec[(n - j - 1) - i + 2000] += a[i][j];
    }
  }
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) {
      res[i][j] = m[j - i + 2000] + sec[(n - j - 1) - i + 2000] - a[i][j];
    }
  }
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) {
      if ((i + j) % 2 == 0) {
        if (res[i][j] > res[wmax.first][wmax.second]) {
          wmax.first = i;
          wmax.second = j;
        }
      } else {
        if (res[i][j] > res[bmax.first][bmax.second]) {
          bmax.first = i;
          bmax.second = j;
        }
      }
    }
  }
  cout << res[bmax.first][bmax.second] + res[wmax.first][wmax.second] << '\n';
  cout << bmax.first + 1 << ' ' << bmax.second + 1 << ' ' << wmax.first + 1
       << ' ' << wmax.second + 1 << '\n';
  return 0;
}
