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 T, typename T2>
ostream &operator<<(ostream &s, const pair<T, T2> &p) {
  return s << "(" << p.first << "," << p.second << ")";
}
template <typename T, typename T2>
istream &operator>>(istream &s, const pair<T, T2> &p) {
  return s >> p.first >> p.second;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  for (auto it : v) s << it << " ";
  return s;
}
template <typename T>
void read_input(T beg, T end) {
  while (beg != end) cin >> *beg++;
}
struct debugger {
  template <typename T>
  debugger &operator,(const T &v) {
    cerr << v << ", ";
    return *this;
  }
} dbg;
int ar[2005][2005];
long long p[2005][2005];
long long max_p[2] = {-1, -1};
int m_x[2], m_y[2];
int n;
void left2right(int x, int y) {
  long long sum = 0;
  int temp_x = x, temp_y = y;
  for (int i = 0; i < x - y + 1; i++) {
    sum += ar[temp_x][temp_y];
    temp_x--, temp_y++;
  }
  temp_x = x, temp_y = y;
  for (int i = 0; i < x - y + 1; i++) {
    p[temp_x][temp_y] += sum;
    temp_x--, temp_y++;
  }
}
void right2left(int x, int y) {
  long long sum = 0;
  int temp_x = x, temp_y = y;
  for (int i = 0; i < x + y + 2 - n; i++) {
    sum += ar[temp_x][temp_y];
    temp_x--, temp_y--;
  }
  temp_x = x, temp_y = y;
  for (int i = 0; i < x + y + 2 - n; i++) {
    p[temp_x][temp_y] += sum;
    temp_x--, temp_y--;
  }
}
void max_l2r(int x, int y, int rem) {
  int temp_x = x, temp_y = y;
  for (int i = 0; i < x - y + 1; i++) {
    if (p[temp_x][temp_y] - ar[temp_x][temp_y] > max_p[rem]) {
      max_p[rem] = p[temp_x][temp_y] - ar[temp_x][temp_y];
      m_x[rem] = temp_x, m_y[rem] = temp_y;
    }
    temp_x--, temp_y++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) cin >> ar[i][j];
  for (int i = 0; i < n - 1; i++) {
    left2right(i, 0);
    left2right(n - 1, i);
  }
  left2right(n - 1, n - 1);
  for (int i = 0; i < n - 1; i++) {
    right2left(i, n - 1);
    right2left(n - 1, i);
  }
  right2left(n - 1, n - 1);
  for (int i = 0; i < 2 * n - 1; i++) {
    if (i < n)
      max_l2r(i, 0, i % 2);
    else
      max_l2r(n - 1, i - n + 1, i % 2);
  }
  cout << max_p[0] + max_p[1] << '\n'
       << m_x[0] + 1 << ' ' << m_y[0] + 1 << ' ' << m_x[1] + 1 << ' '
       << m_y[1] + 1;
  return 0;
}
