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 pre_a[10000];
long long pre_b[10000];
long long Map[2005][2005];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%lld", &Map[i][j]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      pre_a[i + j] += Map[i][j];
      pre_b[i - j + n] += Map[i][j];
    }
  }
  long long ma1 = 0;
  long long ma2 = 0;
  int x1 = 0, y1 = 0, x2 = 0, y2 = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2) {
        int a = i + j;
        int b = i - j + n;
        long long now = pre_a[a] + pre_b[b] - Map[i][j];
        if (now > ma1) {
          ma1 = now;
          x1 = i;
          y1 = j;
        }
      } else {
        int a = i + j;
        int b = i - j + n;
        long long now = pre_a[a] + pre_b[b] - Map[i][j];
        if (now > ma2) {
          ma2 = now;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << ma1 + ma2 << endl;
  cout << x1 + 1 << ' ' << y1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1 << endl;
  return 0;
}
