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;
const long long NR = 5e3 + 20;
const long long oo = 1e16 + 100;
const long long MOD = 1e9 + 7;
long long n, a[2002][6002];
long long st_jos[2002][6002];
signed main() {
  long long i, j;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; ++i) {
    for (j = 1 + n; j <= n * 2; ++j) {
      cin >> a[i][j];
    }
  }
  for (j = 1; j <= n * 2; ++j) {
    long long sum = 0;
    for (long long i1 = 1, j1 = j; i1 <= n; ++i1, ++j1) {
      sum += a[i1][j1];
    }
    for (long long i1 = 1, j1 = j; i1 <= n; ++i1, ++j1) {
      st_jos[i1][j1] += sum;
    }
  }
  for (j = n * 3; j >= n; --j) {
    long long sum = 0;
    for (long long i1 = 1, j1 = j; i1 <= n; ++i1, --j1) {
      sum += a[i1][j1];
    }
    for (long long i1 = 1, j1 = j; i1 <= n; ++i1, --j1) {
      st_jos[i1][j1] += sum - a[i1][j1];
    }
  }
  long long best1 = -1, best2 = -1, x_1, y_1, x_2, y_2;
  for (i = 1; i <= n; ++i) {
    for (j = n + 1; j <= n * 2; ++j) {
      if ((i + j) % 2 && st_jos[i][j] > best1) {
        best1 = st_jos[i][j];
        x_1 = i;
        y_1 = j - n;
      }
      if ((i + j) % 2 == 0 && st_jos[i][j] > best2) {
        best2 = st_jos[i][j];
        x_2 = i;
        y_2 = j - n;
      }
    }
  }
  cout << best1 + best2 << '\n';
  cout << x_1 << ' ' << y_1 << ' ' << x_2 << ' ' << y_2 << '\n';
  return 0;
}
