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 long long INF = 1e12;
const int MAXN = 1e5 + 1, MODM = 1e9 + 7, MULT1 = 30, MULT2 = 239, LUCKY = 13,
          ELEVEN = 11, MAXM = 101;
struct cl {
  long long val, x, y;
  cl() { val = -1; }
};
long long mas[2001][2001];
long long le[2001][2001], rr[2][5001];
long long ans[2001][2001];
cl f_mx[2][2], s_mx[2][2];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> mas[i][j];
      le[i - min(i, j)][j - min(i, j)] += mas[i][j];
      rr[0][j + i] += mas[i][j];
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      ans[i][j] = -mas[i][j] + le[i - min(i, j)][j - min(i, j)] + rr[0][j + i];
      int i1 = i % 2, j1 = j % 2;
      if (f_mx[i1][j1].val < ans[i][j]) {
        f_mx[i1][j1].val = ans[i][j];
        f_mx[i1][j1].x = i;
        f_mx[i1][j1].y = j;
      }
    }
  long long sum = -1, x1 = 0, y1 = 0, x2 = 0, y2 = 0;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) {
      if (sum < f_mx[i][j].val + f_mx[!i][j].val) {
        sum = f_mx[i][j].val + f_mx[!i][j].val;
        x1 = f_mx[i][j].x;
        y1 = f_mx[i][j].y;
        x2 = f_mx[!i][j].x;
        y2 = f_mx[!i][j].y;
      }
      if (sum < f_mx[i][j].val + f_mx[i][!j].val) {
        sum = f_mx[i][j].val + f_mx[i][!j].val;
        x1 = f_mx[i][j].x;
        y1 = f_mx[i][j].y;
        x2 = f_mx[i][!j].x;
        y2 = f_mx[i][!j].y;
      }
    }
  cout << sum << "\n"
       << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1;
  return 0;
}
