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;
int n;
int bd[2000][2000];
long long v[4000], w[4000];
void calc_v(int x, int y) {
  long long sum = 0;
  while (x < n && y < n) {
    sum += bd[x][y];
    ++x;
    ++y;
  }
  v[x - y + 2000] = sum;
}
void calc_w(int x, int y) {
  long long sum = 0;
  while (x < n && y >= 0) {
    sum += bd[x][y];
    ++x;
    --y;
  }
  w[x + y] = sum;
}
long long solve(int k, int &x, int &y) {
  long long ret = -1;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(n); j++) {
      if ((i + j) % 2 != k) continue;
      long long tmp = v[i - j + 2000] + w[i + j] - bd[i][j];
      if (tmp > ret) {
        ret = tmp;
        x = i + 1;
        y = j + 1;
      }
    }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(n); j++) cin >> bd[i][j];
  for (int i = 0; i < (int)(n); i++) calc_v(0, i);
  for (int i = 0; i < (int)(n); i++) calc_v(i, 0);
  for (int i = 0; i < (int)(n); i++) calc_w(0, i);
  for (int i = 0; i < (int)(n); i++) calc_w(i, n - 1);
  int x1, x2, y1, y2;
  long long ret = solve(0, x1, y1) + solve(1, x2, y2);
  cout << ret << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  return 0;
}
