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 INF = 1e-9;
const int EPS = 1e9 + 7;
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<long long> > a(n, vector<long long>(n));
  vector<long long> b(n * 2), c(n * 2);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cin >> a[i][j];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      b[i - j + n] += a[i][j];
      c[i + j] += a[i][j];
    }
  }
  int i1 = 0, j1 = 0, i2 = 1, j2 = 0;
  long long sum0 = b[n] + c[0] - a[0][0], sum1 = b[n + 1] + c[1] - a[1][0];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i + j) % 2) {
        if (sum0 < b[i - j + n] + c[i + j] - a[i][j]) {
          sum0 = b[i - j + n] + c[i + j] - a[i][j];
          i1 = i;
          j1 = j;
        }
      } else {
        if (sum1 < b[i - j + n] + c[i + j] - a[i][j]) {
          sum1 = b[i - j + n] + c[i + j] - a[i][j];
          i2 = i;
          j2 = j;
        }
      }
    }
  }
  cout << sum0 + sum1 << endl
       << i1 + 1 << " " << j1 + 1 << " " << i2 + 1 << " " << j2 + 1;
  return 0;
}
