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;
bool sortt(pair<int, int> &a, pair<int, int> &b) {
  if (a.first == b.first) {
    return (a.second > b.second);
  }
  return (a.first < b.first);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  long long a[n + 1][n + 1];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
  map<long long, long long> m;
  map<long long, long long> mm;
  long long i = 1;
  long long s = 0;
  for (int j = 1; j <= n; j++) {
    s = a[i][j];
    int y = i + 1;
    int z = j + 1;
    while (y <= n && z <= n && y >= 1 && z >= 1) {
      s = s + a[y][z];
      y++;
      z++;
    }
    m[j - i] = s;
  }
  int j = 1;
  for (int i = 1; i <= n; i++) {
    s = a[i][j];
    int y = i + 1;
    int z = j + 1;
    while (y <= n && z <= n && y >= 1 && z >= 1) {
      s = s + a[y][z];
      y++;
      z++;
    }
    m[j - i] = s;
  }
  i = 1;
  for (int j = 1; j <= n; j++) {
    s = a[i][j];
    int y = i + 1;
    int z = j - 1;
    while (y <= n && z <= n && y >= 1 && z >= 1) {
      s = s + a[y][z];
      y++;
      z--;
    }
    mm[j + i] = s;
  }
  j = n;
  for (int i = 1; i <= n; i++) {
    s = a[i][j];
    int y = i + 1;
    int z = j - 1;
    while (y <= n && z <= n && y >= 1 && z >= 1) {
      s = s + a[y][z];
      y++;
      z--;
    }
    mm[j + i] = s;
  }
  long long m1 = -1;
  long long m2 = -1;
  long long ei, ej, oi, oj;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      a[i][j] = m[j - i] + mm[j + i] - a[i][j];
      if ((i + j) % 2 == 1) {
        if (a[i][j] > m1) {
          m1 = a[i][j];
          oi = i;
          oj = j;
        }
      }
      if ((i + j) % 2 == 0) {
        if (a[i][j] > m2) {
          m2 = a[i][j];
          ei = i;
          ej = j;
        }
      }
    }
  }
  cout << m1 + m2 << "\n";
  cout << ei << " " << ej << " " << oi << " " << oj;
}
