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;
inline const long long getint() {
  long long r = 0, k = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') k = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) r = r * 10 + c - '0';
  return k * r;
}
inline const int max(const int &a, const int &b) { return a > b ? a : b; }
inline const int min(const int &a, const int &b) { return a < b ? a : b; }
const int N = 2005, oo = ~0u >> 1;
long long mp[N][N], lft[N + N], rgh[N + N];
int n;
int main() {
  cin >> n;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (n); ++j) mp[i][j] = getint();
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (n); ++j) {
      lft[i + j] += mp[i][j];
      rgh[n + (i - j)] += mp[i][j];
    }
  int xx = -1, yy = -1, xxx = -1, yyy = -1;
  long long now1 = -oo, now2 = -oo;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (n); ++j) {
      if ((i + j) % 2) {
        if (xx == -1 || (lft[i + j] - mp[i][j] + rgh[n + i - j]) > now1) {
          xx = i;
          yy = j;
          now1 = (lft[i + j] - mp[i][j] + rgh[n + i - j]);
        }
      } else {
        if (xxx == -1 || (lft[i + j] - mp[i][j] + rgh[n + i - j]) > now2) {
          xxx = i;
          yyy = j;
          now2 = (lft[i + j] - mp[i][j] + rgh[n + i - j]);
        }
      }
    }
  cout << now1 + now2 << endl;
  cout << xxx << " " << yyy << " " << xx << " " << yy << endl;
  return 0;
}
