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 int p[2003][2003], r[2003][2003], l[2003][2003];
int n, rgenap, cgenap, rganjil, cganjil;
long long int calc(int col, int row) {
  int mini;
  mini = min(n - col, n - row);
  long long int a = r[col + mini][row + mini];
  mini = min(col - 1, n - row);
  long long int b = l[col - mini][row + mini];
  return a + b - p[col][row];
}
int main(void) {
  scanf("%d", &n);
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= n; i++) scanf("%I64d", &p[i][j]);
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= n; i++) r[i][j] = p[i][j] + r[i - 1][j - 1];
  for (int j = 1; j <= n; j++)
    for (int i = n; i >= 1; i--) l[i][j] = p[i][j] + l[i + 1][j - 1];
  rgenap = cgenap = 1;
  rganjil = 1;
  cganjil = 2;
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= n; i++) {
      if ((i + j) % 2 == 1) {
        if (calc(i, j) > calc(cganjil, rganjil)) {
          cganjil = i;
          rganjil = j;
        }
      } else {
        if (calc(i, j) > calc(cgenap, rgenap)) {
          cgenap = i;
          rgenap = j;
        }
      }
    }
  printf("%I64d\n", calc(cganjil, rganjil) + calc(cgenap, rgenap));
  printf("%d %d %d %d\n", rganjil, cganjil, rgenap, cgenap);
}
