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;
vector<vector<int64_t>> board, vis, diag1, diag2;
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int64_t n;
  cin >> n;
  for (int64_t i = 0; i < n; i++) {
    vector<int64_t> temp;
    vector<int64_t> zero;
    for (int64_t j = 0; j < n; j++) {
      int64_t t;
      cin >> t;
      temp.push_back(t);
      zero.push_back(0);
    }
    board.push_back(temp);
    vis.push_back(zero);
    diag1.push_back(zero);
    diag2.push_back(zero);
  }
  for (int64_t i = 0; i < n; i++) {
    for (int64_t j = 0; j < n; j++) {
      if (vis[i][j]) continue;
      int64_t t1 = i, t2 = j, s = 0;
      while (1) {
        vis[t1][t2] = 1;
        s += board[t1][t2];
        t1 -= -1;
        t2 -= -1;
        if (t1 >= n || t2 >= n) break;
      }
      t1 = i, t2 = j;
      while (1) {
        diag1[t1][t2] = s;
        t1 -= -1;
        t2 -= -1;
        if (t1 >= n || t2 >= n) break;
      }
    }
  }
  for (int64_t i = 0; i < n; i++)
    for (int64_t j = 0; j < n; j++) vis[i][j] = 0;
  for (int64_t i = 0; i < n; i++) {
    for (int64_t j = 0; j < n; j++) {
      if (vis[i][j]) continue;
      int64_t t1 = i, t2 = j, s = 0;
      while (1) {
        vis[t1][t2] = 1;
        s += board[t1][t2];
        t1 -= -1;
        t2 += -1;
        if (t1 >= n || t2 < 0) break;
      }
      t1 = i, t2 = j;
      while (1) {
        diag2[t1][t2] = s;
        t1 -= -1;
        t2 += -1;
        if (t1 >= n || t2 < 0) break;
      }
    }
  }
  int64_t m1 = 0, m2 = 0, x1 = 0, x2 = 0, y1 = 0, y2 = 0;
  for (int64_t i = 0; i < n; i++) {
    for (int64_t j = 0; j < n; j++) {
      int64_t tmp = diag1[i][j] + diag2[i][j] - board[i][j];
      if ((i + j) % 2) {
        if (tmp >= m1) {
          m1 = tmp;
          x1 = i + 1;
          y1 = j + 1;
        }
      } else {
        if (tmp >= m2) {
          m2 = tmp;
          x2 = i + 1;
          y2 = j + 1;
        }
      }
    }
  }
  cout << m1 + m2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
}
