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;
#pragma comment(linker, "/STACK:128000000")
int n;
vector<vector<int> > a;
vector<long long> d1, d2;
long long b1 = -1, b2 = -1;
pair<int, int> cell1(-1, -1), cell2(-1, -1);
inline void init() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    a[i].resize(n);
    for (int j = 0; j < n; ++j) scanf("%d", &a[i][j]);
  }
  d1.assign(3 * n, 0LL);
  d2.assign(3 * n, 0LL);
}
int main() {
  init();
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      d1[i + j] += (long long)a[i][j];
      d2[i - j + n] += (long long)a[i][j];
    }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      long long cur = d1[i + j] + d2[i - j + n] - (long long)a[i][j];
      int color = (i + j) & 1;
      if (color) {
        if (b1 < cur) {
          b1 = cur;
          cell1 = pair<int, int>(i, j);
        }
      } else {
        if (b2 < cur) {
          b2 = cur;
          cell2 = pair<int, int>(i, j);
        }
      }
    }
  cout << b1 + b2 << endl;
  cout << cell1.first + 1 << " " << cell1.second + 1 << " " << cell2.first + 1
       << " " << cell2.second + 1 << endl;
  return 0;
}
