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 N = 2005;
long long OFFSET = N;
long long one[5 * N], two[5 * N];
long long dp[N][N];
long long a[N][N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      one[OFFSET + i + j] += a[i][j];
      two[OFFSET + i - j] += a[i][j];
    }
  }
  long long ans = -1;
  long long ax, ay, bx, by;
  long long aone = -1, bone = -1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      dp[i][j] = one[OFFSET + i + j] + two[OFFSET + i - j] - a[i][j];
      if ((i + j) % 2 == 0) {
        if (dp[i][j] > aone) {
          aone = dp[i][j];
          ax = i;
          ay = j;
        }
      } else {
        if (dp[i][j] > bone) {
          bone = dp[i][j];
          bx = i;
          by = j;
        }
      }
    }
  }
  cout << aone + bone << "\n";
  cout << ax << " " << ay << " " << bx << " " << by << "\n";
}
