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;
int main() {
  long long int n;
  scanf("%lld", &n);
  long long int a[n][n];
  map<long long int, long long int> dial, diar;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      scanf("%lld", &a[i][j]);
      dial[i - j] += a[i][j];
      diar[i + j] += a[i][j];
    }
  }
  long long int ma = -1;
  long long int x = -1, y = -1;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      if (ma < (dial[i - j] + diar[i + j] - a[i][j])) {
        ma = (dial[i - j] + diar[i + j] - a[i][j]);
        x = i;
        y = j;
      }
    }
  }
  long long int ma2 = -1;
  long long int x1 = -1, y1 = -1;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      if (ma2 < (dial[i - j] + diar[i + j] - a[i][j]) && (i != x || j != y) &&
          ((i + j) % 2 != (x + y) % 2) && ((i - j) % 2 != (x - y) % 2)) {
        ma2 = (dial[i - j] + diar[i + j] - a[i][j]);
        x1 = i;
        y1 = j;
      }
    }
  }
  if (x1 == -1 || y1 == -1) {
    x1 = x;
    y1 = y + 1;
  }
  printf("%lld\n%lld %lld %lld %lld", ma2 + ma, x + 1, y + 1, x1 + 1, y1 + 1);
  return 0;
}
