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;
long long m[2000][2000];
long long r[2][5000];
long long res;
int n;
int x[2], y[2];
void find_s() {
  long long max[2];
  int p, f, s;
  max[0] = max[1] = -1;
  p = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      p = (j + i) & 1;
      if (i >= j)
        f = i - j;
      else
        f = n + (j - i);
      if (n - i > j)
        s = i + j;
      else
        s = j + i + 1;
      if (max[p] < r[0][f] + r[1][s] - m[i][j]) {
        max[p] = r[0][f] + r[1][s] - m[i][j];
        x[p] = i;
        y[p] = j;
      }
    }
  res = max[0] + max[1];
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf("%d", &(m[j][i]));
  for (int i = 0; i < n; i++) {
    r[0][i] = 0;
    for (int j = 0; i + j < n; j++) r[0][i] += m[i + j][j];
  }
  for (int i = 0; i < n; i++) {
    r[0][n + i] = 0;
    for (int j = 0; j + i < n; j++) r[0][n + i] += m[j][j + i];
  }
  for (int i = 0; i < n; i++) {
    r[1][i] = 0;
    for (int j = 0; i - j >= 0; j++) r[1][i] += m[i - j][j];
  }
  for (int i = 0; i < n; i++) {
    r[1][n + i] = 0;
    for (int j = 0; i + j < n; j++) r[1][n + i] += m[n - j][i + j];
  }
  for (int i = 0; i < n; i++) {
    r[1][n + i] = 0;
    for (int j = 0; j < n - i; j++) r[1][n + i] += m[n - j - 1][i + j];
  }
  find_s();
  cout << res << endl;
  cout << y[0] + 1 << " " << x[0] + 1 << " " << y[1] + 1 << " " << x[1] + 1
       << endl;
  return 0;
}
