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 gcd(int a, int b) {
  if (a < b) {
    a ^= b;
    b ^= a;
    a ^= b;
  }
  return (a - b) ? gcd(a - b, b) : a;
}
void solve() {
  unsigned long long n, amax = 0, bmax = 0, t, tmp;
  int x, y, ax = 0, ay = 0, bx = 0, by = 1;
  cin >> n;
  vector<vector<unsigned long long> > a;
  vector<unsigned long long> emp(n), left(2 * n - 1), right(2 * n - 1);
  for (int i = 0; i < n; i++) a.push_back(emp);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      left[n - i + j - 1] += a[i][j];
      right[i + j] += a[i][j];
    }
  for (int i = 0; i < 2 * n - 1; i++)
    for (int j = 0; j < 2 * n - 1; j++) {
      if ((i + j + n) % 2 == 0) continue;
      x = (n + i - j - 1) / 2, y = (-n + i + j + 1) / 2;
      if (x < 0 || x >= n || y < 0 || y >= n) continue;
      t = a[x][y];
      tmp = left[j] + right[i] - t;
      if ((i % 2)) {
        if (tmp > amax) {
          amax = tmp;
          ax = x;
          ay = y;
        }
      } else {
        if (tmp > bmax) {
          bmax = tmp;
          bx = x;
          by = y;
        }
      }
    }
  cout << amax + bmax << endl;
  cout << ax + 1 << " " << ay + 1 << " " << bx + 1 << " " << by + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
