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 ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 2001;
int n, a[N][N];
ll dia[2][2 * N], mx[2];
pair<int, int> po[2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (n); ++j) {
      cin >> a[i][j];
      dia[0][i + j - 1] += a[i][j];
      dia[1][j - i + n] += a[i][j];
    }
  mx[0] = mx[1] = -1;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (n); ++j) {
      if (mx[((i + j) % 2)] < dia[0][i + j - 1] + dia[1][j - i + n] - a[i][j]) {
        mx[((i + j) % 2)] = dia[0][i + j - 1] + dia[1][j - i + n] - a[i][j];
        po[((i + j) % 2)] = {i, j};
      }
    }
  cout << mx[0] + mx[1] << '\n'
       << po[0].first << ' ' << po[0].second << ' ' << po[1].first << ' '
       << po[1].second;
  return 0;
}
