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;
bool sortmahstyle(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first && a.second < b.second) return true;
  return false;
}
long long int b[1000006], c[1000006];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  long long int n;
  cin >> n;
  long long int a[n + 1][n + 1];
  long long int x1, y1, x2, y2;
  for (long long int i = 1; i < n + 1; i++)
    for (long long int j = 1; j < n + 1; j++) {
      cin >> a[i][j];
      b[i + j] += a[i][j];
      c[i - j + n] += a[i][j];
    }
  long long int ans1, ans2, ans3, ans4;
  ans1 = ans2 = ans3 = ans4 = LLONG_MIN;
  long long int A[4], B[4];
  for (long long int i = 1; i < n + 1; i++)
    for (long long int j = 1; j < n + 1; j++) {
      if ((i + j) % 2 != 0) {
        if (b[i + j] + c[i - j + n] - a[i][j] > ans1) {
          ans1 = b[i + j] + c[i - j + n] - a[i][j];
          A[0] = i, A[1] = j;
        }
      }
      if ((i + j) % 2 == 0) {
        if (b[i + j] + c[i - j + n] - a[i][j] > ans3) {
          ans3 = b[i + j] + c[i - j + n] - a[i][j];
          A[2] = i, A[3] = j;
        }
      }
    }
  cout << ans1 + ans3 << "\n";
  for (long long int i = 0; i < 4; i++) cout << A[i] << " ";
  return 0;
}
