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 n;
long long tbl[2001][2001];
long long dia1[4001];
long long dia2[4001];
long long max_odd, max_even;
int x1, yc1, x2, y2;
int main() {
  ios_base::sync_with_stdio(0);
  memset(dia1, 0, sizeof dia1);
  memset(dia2, 0, sizeof dia2);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cin >> tbl[i][j];
      int idx1 = n + (j - i);
      int idx2 = i + j;
      dia1[idx1] += tbl[i][j];
      dia2[idx2] += tbl[i][j];
    }
  }
  max_odd = -1;
  x1 = -1;
  yc1 = -1;
  max_even = -1;
  x2 = -1;
  y2 = -1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      int idx1 = n + (j - i);
      int idx2 = i + j;
      long long val = dia1[idx1] + dia2[idx2] - tbl[i][j];
      if ((i + j) % 2) {
        if (max_odd < val) {
          max_odd = val;
          x1 = i;
          yc1 = j;
        }
      } else {
        if (max_even < val) {
          max_even = val;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << max_even + max_odd << endl;
  cout << x1 + 1 << ' ' << yc1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1 << endl;
  return 0;
}
