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 int a[2009][2009];
int n;
map<int, long long int> sum, diff;
pair<long long int, pair<int, int> > b[4000009];
int main() {
  memset(b, 0, sizeof(b));
  scanf("%d", &n);
  for (int i = int(1); i <= int(n); i++) {
    for (int j = int(1); j <= int(n); j++) {
      scanf("%lld", &a[i][j]);
      diff[(i - j)] += a[i][j];
      sum[i + j] += a[i][j];
    }
  }
  int co = 0;
  for (int i = int(1); i <= int(n); i++) {
    for (int j = int(1); j <= int(n); j++) {
      b[co++] = make_pair(sum[i + j] + diff[i - j] - a[i][j], make_pair(i, j));
    }
  }
  sort(b, b + co);
  reverse(b, b + co);
  int q = n * n;
  long long int ans = b[0].first;
  pair<int, int> fir = b[0].second;
  pair<int, int> sec;
  for (int i = int(1); i <= int(q - 1); i++) {
    if ((abs(b[i].second.first - fir.first) +
         abs(b[i].second.second - fir.second)) %
            2 ==
        1) {
      ans += b[i].first;
      sec = b[i].second;
      break;
    }
  }
  printf("%lld\n", ans);
  printf("%d %d %d %d\n", fir.first, fir.second, sec.first, sec.second);
  return 0;
}
