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 scan() {
  char c;
  while (c = getchar(), (c < '0' || c > '9') && (c != '-'))
    ;
  bool flag = (c == '-');
  if (flag) c = getchar();
  int x = 0;
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - 48;
    c = getchar();
  }
  return flag ? -x : x;
}
const int maxn = 2345;
int a[maxn][maxn];
long long l[maxn << 1], r[maxn << 1];
int n;
int main() {
  n = scan();
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (n); ++j) {
      a[i][j] = scan();
      l[i + j] += a[i][j];
      r[i - j + n] += a[i][j];
    }
  }
  pair<int, int> x, y;
  long long t1 = -1, t2 = -1;
  long long tmp;
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (n); ++j) {
      if ((i + j) & 1) {
        tmp = l[i + j] + r[i - j + n] - a[i][j];
        if (tmp > t1) {
          t1 = tmp;
          x.first = i, x.second = j;
        }
      } else {
        tmp = l[i + j] + r[i - j + n] - a[i][j];
        if (tmp > t2) {
          t2 = tmp;
          y.first = i, y.second = j;
        }
      }
    }
  }
  cout << t1 + t2 << endl;
  cout << x.first << " " << x.second << " " << y.first << " " << y.second
       << endl;
  return 0;
}
