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;
int ar[2005][2005];
long long ans;
map<int, long long> mp, _mp;
int main() {
  int n = 2000;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%d", &ar[i][j]);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      mp[i - j] += ar[i][j];
      _mp[i + j] += ar[i][j];
    }
  }
  int x, y, x1 = 0, x2 = 0, y1 = 0, y2 = 0;
  long long mx = -1, _mx = 0, MX = -1, _MX = -1;
  bool fl = true;
  int i = 0, j = 0;
  for (int k = 1; k <= (n * n); k++) {
    i = ((k - 1) / n) + 1;
    j = (k % n) + 1;
    if ((i % 2) == 0) j = n - j + 1;
    if (fl) {
      fl = false;
      if ((mp[i - j] + _mp[i + j] - ar[i][j]) > MX) {
        MX = (mp[i - j] + _mp[i + j] - ar[i][j]);
        x1 = i;
        y1 = j;
      }
    } else {
      fl = true;
      if ((mp[i - j] + _mp[i + j] - ar[i][j]) > _MX) {
        _MX = (mp[i - j] + _mp[i + j] - ar[i][j]);
        x2 = i;
        y2 = j;
      }
    }
  }
  cout << MX + _MX << endl << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  return 0;
}
