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 = 2000;
long long a[maxn + 5][maxn + 5], f[4][maxn + 5][maxn + 5];
int n;
long long ans[2], ans2[2][2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) scanf("%lld", &a[i][j]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) f[0][i][j] = f[0][i - 1][j - 1] + a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = n; j >= 1; j--) f[1][i][j] = f[1][i - 1][j + 1] + a[i][j];
  for (int i = n; i >= 1; i--)
    for (int j = 1; j <= n; j++) f[2][i][j] = f[2][i + 1][j - 1] + a[i][j];
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--) f[3][i][j] = f[3][i + 1][j + 1] + a[i][j];
  ans[0] = ans[1] = -1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int id = (i + j) % 2;
      long long tmp = 0;
      for (int k = 0; k < 4; k++) tmp += f[k][i][j];
      tmp -= 3 * a[i][j];
      if (ans[id] < tmp) {
        ans[id] = tmp;
        ans2[id][0] = i;
        ans2[id][1] = j;
      }
    }
  printf("%lld\n", ans[0] + ans[1]);
  printf("%lld %lld %lld %lld\n", ans2[0][0], ans2[0][1], ans2[1][0],
         ans2[1][1]);
  return 0;
}
