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 double pi = acos(-1.0);
long long chess[5][2004][2004];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int N, r1, r2, c1, c2;
  cin >> N;
  for (int i = 1; i < N + 1; i += 1)
    for (int j = 1; j < N + 1; j += 1)
      cin >> chess[0][i][j],
          chess[1][i][j] = chess[0][i][j] + chess[1][i - 1][j - 1],
          chess[2][i][j] = chess[0][i][j] + chess[2][i - 1][j + 1];
  for (int i = N; i >= 1; i -= 1)
    for (int j = N; j >= 1; j -= 1)
      chess[3][i][j] = chess[0][i][j] + chess[3][i + 1][j + 1],
      chess[4][i][j] = chess[0][i][j] + chess[4][i + 1][j - 1];
  long long ec = -1, oc = -1, t;
  for (int i = 1; i < N + 1; i += 1)
    for (int j = 1; j < N + 1; j += 1) {
      t = chess[0][i][j] + chess[1][i - 1][j - 1] + chess[2][i - 1][j + 1] +
          chess[3][i + 1][j + 1] + chess[4][i + 1][j - 1];
      if (i + j & 1 && oc < t)
        r1 = i, c1 = j, oc = t;
      else if ((i + j & 1) == 0 && ec < t)
        r2 = i, c2 = j, ec = t;
    }
  cout << oc + ec << '\n' << r1 << ' ' << c1 << ' ' << r2 << ' ' << c2 << '\n';
}
