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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  long long int a[n][n];
  map<long long int, long long int> sum;
  map<long long int, long long int> dif;
  long long int ans[n][n][2];
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) {
      cin >> a[i][j];
    }
  }
  long long int s, d;
  long long int b = 0, w = 0, x, y;
  long long int x1 = 0, y1 = 0, x2 = 0, y2 = 1;
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) {
      s = i + j;
      x = i;
      y = j;
      d = j - i;
      if (!dif[d]) {
        dif[d] += a[x][y];
        while (x >= 0 && x < n - 1 && y >= 0 && y < n - 1) {
          x++;
          y++;
          dif[d] += a[x][y];
        }
      }
      x = i;
      y = j;
      if (!sum[s]) {
        sum[s] += a[x][y];
        while (x >= 0 && x < n - 1 && y > 0 && y < n) {
          x++;
          y--;
          sum[s] += a[x][y];
        }
      }
      if (s & 1) {
        ans[i][j][0] = dif[d] + sum[s] - a[i][j];
        if (b < ans[i][j][0]) {
          b = ans[i][j][0];
          x1 = i;
          y1 = j;
        }
      } else {
        ans[i][j][1] = dif[d] + sum[s] - a[i][j];
        if (w < ans[i][j][1]) {
          w = ans[i][j][1];
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << b + w << endl;
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1;
  return 0;
}
