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 <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &a) {
  return out << "(" << a.first << "," << a.second << ")";
}
template <class A>
ostream &operator<<(ostream &out, const vector<A> &a) {
  for (const A &it : a) out << it << " ";
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  return in >> a.first >> a.second;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &i : a) in >> i;
  return in;
}
ifstream cinn("in.txt");
ofstream coutt("out.txt");
long long poww(const long long &a, long long b,
               const long long &m = 1000000007) {
  if (b == 0) return 1;
  long long first = poww(a, b / 2, m);
  first = first * first % m;
  if (b & 1) first = first * a % m;
  return first;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long rint(long long l, long long r) { return rand() % (r - l + 1) + l; }
const long long N = 2005;
long long grid[N][N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j, first;
  map<long long, long long> mp, mp1;
  long long n;
  cin >> n;
  long long a, b;
  long long e = -1, o = -1;
  pair<long long, long long> eve, odd;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> grid[i][j];
      a = i + j;
      b = i - j;
      mp[a] += grid[i][j];
      mp1[b] += grid[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      long long sum = 0;
      a = i + j;
      b = i - j;
      sum = mp[a] + mp1[b] - grid[i][j];
      if (a % 2 == 0) {
        if (sum > e) {
          e = sum;
          eve = {i, j};
        }
      } else {
        if (sum > o) {
          o = sum;
          odd = {i, j};
        }
      }
    }
  }
  cout << e + o << '\n';
  cout << eve.first + 1 << " " << eve.second + 1 << " " << odd.first + 1 << " "
       << odd.second + 1 << '\n';
  return 0;
}
