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 = 1e5 + 5;
const int Mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e = exp(1);
const double PI = acos(-1);
const double ERR = 1e-10;
const int nmax = 2005;
long long d1[2 * nmax], d2[2 * nmax], sol[2];
pair<int, int> v[2];
int arr[nmax][nmax];
inline void update(const int c, const int i, const int j, const long long val) {
  if (val > sol[c]) {
    sol[c] = val;
    v[c].first = i;
    v[c].second = j;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  sol[0] = sol[1] = -1;
  for (int i = 1; i < n + 1; i++)
    for (int j = 1; j < n + 1; j++) {
      int x;
      cin >> arr[i][j];
      d1[i + j] += arr[i][j];
      d2[i - j + n] += arr[i][j];
    }
  for (int i = 1; i < n + 1; i++)
    for (int j = 1; j < n + 1; j++) {
      update((i + j) & 1, i, j, d1[i + j] + d2[i - j + n] - arr[i][j]);
    }
  if (v[0] > v[1]) swap(v[0], v[1]);
  cout << sol[0] + sol[1] << "\n";
  cout << v[0].first << " " << v[0].second << " " << v[1].first << " "
       << v[1].second << "\n";
  return 0;
}
