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;
long long cost[2001][2001];
long long a[2001][2001];
signed main() {
  ios_base::sync_with_stdio(false);
  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];
      cost[i][j] = -a[i][j];
    }
  for (long long j = 1; j <= n; j++) {
    long long res = 0;
    for (long long i = 0; j + i <= n; i++) res += a[i + 1][j + i];
    for (long long i = 0; j + i <= n; i++) cost[i + 1][j + i] += res;
  }
  for (long long i = 2; i <= n; i++) {
    long long res = 0;
    for (long long j = i; j <= n; j++) res += a[j][j - i + 1];
    for (long long j = i; j <= n; j++) cost[j][j - i + 1] += res;
  }
  for (long long j = n; j >= 1; j--) {
    long long res = 0;
    for (long long i = 0; j - i >= 1; i++) res += a[i + 1][j - i];
    for (long long i = 0; j - i >= 1; i++) cost[i + 1][j - i] += res;
  }
  for (long long i = 2; i <= n; i++) {
    long long res = 0;
    for (long long j = i; j <= n; j++) res += a[j][n - j + i];
    for (long long j = i; j <= n; j++) cost[j][n - j + i] += res;
  }
  long long resodd = -1;
  long long reseven = -1;
  long long xodd, yodd, xeven, yeven;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      if ((i + j) & 1 && cost[i][j] > resodd) {
        resodd = cost[i][j];
        xodd = i;
        yodd = j;
      }
      if ((i + j) % 2 == 0 && cost[i][j] > reseven) {
        reseven = cost[i][j];
        xeven = i;
        yeven = j;
      }
    }
  }
  cout << reseven + resodd << '\n';
  cout << xodd << " " << yodd << " " << xeven << " " << yeven << '\n';
  return 0;
}
