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 T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  os << "]";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, set<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <class T>
ostream& operator<<(ostream& os, multiset<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, unordered_map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
const long long N = 2010, M = 1e9 + 7;
long long n, a[N][N], dp1[N][N], dp2[N][N], dp3[N][N];
void solve() {
  cin >> n;
  for (long long i = (0); i <= n - 1; i++)
    for (long long j = (0); j <= n - 1; j++) cin >> a[i][j];
  for (long long i = (0); i <= n - 1; i++)
    dp1[i][0] = a[i][0], dp1[0][i] = a[0][i], dp2[i][n - 1] = a[i][n - 1],
    dp2[0][n - i - 1] = a[0][n - i - 1];
  for (long long i = (1); i <= n - 1; i++)
    for (long long j = (1); j <= n - 1; j++)
      dp1[i][j] = a[i][j] + dp1[i - 1][j - 1];
  for (long long i = (1); i <= n - 1; i++)
    for (long long j = (0); j <= n - 2; j++)
      dp2[i][j] = a[i][j] + dp2[i - 1][j + 1];
  for (long long i = (0); i <= n - 1; i++)
    for (long long j = (0); j <= n - 1; j++) {
      long long i1 = i + min(n - i - 1, n - j - 1);
      long long j1 = j + min(n - i - 1, n - j - 1);
      long long i2 = i + min(n - i - 1, j);
      long long j2 = j - min(n - i - 1, j);
      dp3[i][j] = dp1[i1][j1] + dp2[i2][j2] - a[i][j];
    }
  long long ans1 = LLONG_MIN, ans2 = LLONG_MIN, posx1, posy1, posx2, posy2;
  for (long long i = (0); i <= n - 1; i++)
    for (long long j = (0); j <= n - 1; j++) {
      if ((i + j) % 2 == 0 && dp3[i][j] > ans1) {
        ans1 = max(ans1, dp3[i][j]);
        posx1 = i + 1;
        posy1 = j + 1;
      } else if ((i + j) % 2 == 1 && dp3[i][j] > ans2) {
        ans2 = max(ans2, dp3[i][j]);
        posx2 = i + 1;
        posy2 = j + 1;
      }
    }
  cout << ans1 + ans2 << "\n";
  cout << posx1 << ' ' << posy1 << ' ' << posx2 << " " << posy2 << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  solve();
  return 0;
}
