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;
const int oo = 1e9 + 9;
const long long inf = 1e18 + 18;
const int max6 = 1e6 + 6;
const int modx = 1e9 + 123;
const int mody = 997;
const int base = 137;
int a[2005][2005];
long long f[2005][2005];
long long g[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) cin >> a[i][j];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) f[i][j] = f[i - 1][j - 1] + a[i][j];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) g[i][j] = g[i - 1][j + 1] + a[i][j];
  pair<long long, pair<int, int> > max_odd = {0, {0, 0}};
  pair<long long, pair<int, int> > max_even = {0, {0, 0}};
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      int one = min(n - i, n - j);
      int two = min(n - i, j - 1);
      long long cur = f[i + one][j + one] + g[i + two][j - two] - a[i][j];
      if ((i + j) % 2)
        max_odd = max(max_odd, {cur, {i, j}});
      else
        max_even = max(max_even, {cur, {i, j}});
    }
  cout << max_odd.first + max_even.first << "\n";
  cout << max_odd.second.first << " " << max_odd.second.second << " ";
  cout << max_even.second.first << " " << max_even.second.second << " ";
}
