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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long maxn = (long long)2010;
const long double EPS = 1e-9;
const long long INF = (long long)1e18 + 18;
const long long mod = (long long)1e9 + 7;
long long n, a[maxn][maxn], dp1[2 * maxn], dp2[2 * maxn], s1 = -INF, s2 = -INF;
pair<long long, long long> p1, p2;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cin >> a[i][j];
      dp1[i + j] += a[i][j];
      dp2[i - j + n - 1] += a[i][j];
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      if (i % 2 == j % 2) {
        if (dp1[i + j] + dp2[i - j + n - 1] - a[i][j] > s1) {
          s1 = dp1[i + j] + dp2[i - j + n - 1] - a[i][j];
          p1 = {i + 1, j + 1};
        }
      } else {
        if (dp1[i + j] + dp2[i - j + n - 1] - a[i][j] > s2) {
          s2 = dp1[i + j] + dp2[i - j + n - 1] - a[i][j];
          p2 = {i + 1, j + 1};
        }
      }
    }
  }
  cout << s1 + s2 << endl;
  cout << p1.first << " " << p1.second << " " << p2.first << " " << p2.second
       << endl;
  return 0;
}
