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 N = 2010;
long long n, m, i, j, k, q, s, w, v, ans, x, y, ans_2, xx, yy, xxx, yyy;
long long a[N][N], b[N][N], c[N][N];
int main() {
  cin >> n;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) scanf("%d", &a[i][j]);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      b[i][j] = b[i - 1][j - 1] + a[i][j];
      c[i][j] = c[i - 1][j + 1] + a[i][j];
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        x = i;
        y = j;
        k = min(n - i, n - j);
        s = min(n - i, j - 1);
        if (ans < b[x + k][y + k] + c[x + s][y - s] - a[i][j]) {
          ans = b[x + k][y + k] + c[x + s][y - s] - a[i][j];
          xx = x;
          yy = y;
        }
      } else {
        x = i;
        y = j;
        k = min(n - i, n - j);
        s = min(n - i, j - 1);
        if (ans_2 < c[s + x][y - s] + b[x + k][y + k] - a[i][j]) {
          ans_2 = c[s + x][y - s] + b[x + k][y + k] - a[i][j];
          xxx = x;
          yyy = y;
        }
      }
    }
  }
  cout << ans_2 + ans << endl;
  if (ans_2 + ans == 0) {
    cout << "1 1 1 2" << endl;
    return 0;
  }
  cout << xx << " " << yy << " " << xxx << " " << yyy << endl;
  return 0;
}
