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 int N = 2002;
int a[N][N];
long long b[N + N], c[N + N];
int n;
int main() {
  int i, j;
  long long m1 = 0, m2 = 0;
  int mx1 = 1, my1 = 1, mx2 = 1, my2 = 2;
  long long k;
  scanf("%d", &n);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      scanf("%d", &a[i][j]);
      b[i + j] = b[i + j] + a[i][j];
      c[i - j + n - 1] = c[i - j + n - 1] + a[i][j];
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      k = b[i + j] + c[i - j + n - 1] - a[i][j];
      if ((i + j) % 2 == 0 && k > m1) {
        m1 = k;
        mx1 = i + 1;
        my1 = j + 1;
      }
      if ((i + j) % 2 == 1 && k > m2) {
        m2 = k;
        mx2 = i + 1;
        my2 = j + 1;
      }
    }
  cout << m1 + m2 << endl;
  cout << mx1 << " " << my1 << " " << mx2 << " " << my2 << endl;
  return 0;
}
