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 a[2002][2002], b[2002][2002], c[2002][2002];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long tmp;
      scanf("%I64d", &tmp);
      a[i][j] = tmp;
      b[i][j] = b[i - 1][j - 1] + tmp;
      c[i][j] = c[i - 1][j + 1] + tmp;
    }
  }
  long long ans0 = -1, ans1 = -1;
  int i1 = 0, i2 = 0, j1 = 0, j2 = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long tmp1, tmp2;
      tmp1 = b[i - 1][j - 1];
      if (i == j) {
        tmp1 += b[n][n] - b[i][j];
      } else if (i > j) {
        int kol = n - (i - j);
        tmp1 += b[n][kol] - b[i][j];
      } else if (i < j) {
        int bar = n - (j - i);
        tmp1 += b[bar][n] - b[i][j];
      }
      tmp2 = c[i - 1][j + 1];
      if (i + j == n + 1) {
        tmp2 += c[n][1] - c[i][j];
      } else if (i + j < n + 1) {
        int bar = n - (n + 1 - (i + j));
        tmp2 += c[bar][1] - c[i][j];
      } else if (i + j > n + 1) {
        int kol = 1 + (i + j - (n + 1));
        tmp2 += c[n][kol] - c[i][j];
      }
      if ((i + j) % 2 == 0) {
        if (tmp1 + tmp2 + a[i][j] > ans0) {
          ans0 = tmp1 + tmp2 + a[i][j];
          i1 = i;
          j1 = j;
        }
      } else if ((i + j) % 2 == 1) {
        if (tmp1 + tmp2 + a[i][j] > ans1) {
          ans1 = tmp1 + tmp2 + a[i][j];
          i2 = i;
          j2 = j;
        }
      }
    }
  }
  cout << ans0 + ans1 << endl;
  printf("%d %d %d %d\n", i1, j1, i2, j2);
  return 0;
}
