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 int MAXN = 2005;
int grid[MAXN][MAXN];
long long diag_b[MAXN][MAXN];
long long diag_f[MAXN][MAXN];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf("%d", &grid[i][j]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      diag_b[i][j] = grid[i][j] + (i > 0 && j > 0 ? diag_b[i - 1][j - 1] : 0);
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      diag_f[i][j] =
          grid[i][j] + (i > 0 && j < n - 1 ? diag_f[i - 1][j + 1] : 0);
    }
  long long maxv = -1, maxv2 = -1;
  int x1 = -1, y1 = -1, x2 = -1, y2 = -1;
  for (int i = 0; i < n; i++) {
    for (int j = (i & 1) ? 1 : 0; j < n; j += 2) {
      long long dist, nowb_sum = 0, nowf_sum = 0;
      dist = min(n - i - 1, n - j - 1);
      nowb_sum = diag_b[i + dist][j + dist];
      dist = min(n - 1 - i, j);
      nowf_sum = diag_f[i + dist][j - dist];
      if (nowf_sum + nowb_sum - grid[i][j] > maxv) {
        maxv = nowf_sum + nowb_sum - grid[i][j];
        x1 = i + 1, y1 = j + 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = (i & 1) ? 0 : 1; j < n; j += 2) {
      long long dist, nowb_sum = 0, nowf_sum = 0;
      dist = min(n - i - 1, n - j - 1);
      nowb_sum = diag_b[i + dist][j + dist];
      dist = min(n - 1 - i, j);
      nowf_sum = diag_f[i + dist][j - dist];
      if (nowf_sum + nowb_sum - grid[i][j] > maxv2) {
        maxv2 = nowf_sum + nowb_sum - grid[i][j];
        x2 = i + 1, y2 = j + 1;
      }
    }
  }
  cout << maxv + maxv2 << '\n';
  cout << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2 << '\n';
  return 0;
}
