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;
constexpr size_t kMaxN{2100};
int n;
long long field[kMaxN + 10][kMaxN + 10];
bool color[kMaxN + 10][kMaxN + 10];
int left_diag[kMaxN + 10][kMaxN + 10];
int right_diag[kMaxN + 10][kMaxN + 10];
long long left_diag_sum[3 * kMaxN + 10];
long long right_diag_sum[3 * kMaxN + 10];
inline bool valid(int i, int j) { return i >= 0 && i < n && j >= 0 && j < n; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) cin >> field[i][j];
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      if (i % 2 == 0)
        color[i][j] = j % 2;
      else
        color[i][j] = (j + 1) % 2;
    }
  for (int i = 0; i < n; ++i) {
    int ti = i;
    int tj = i;
    while (valid(ti + 1, tj - 1)) {
      ++ti;
      --tj;
    }
    while (valid(ti, tj)) {
      left_diag[ti][tj] = 2 * i + 1;
      --ti;
      ++tj;
    }
    ti = i;
    tj = i + 1;
    if (!valid(ti, tj)) continue;
    while (valid(ti + 1, tj - 1)) {
      ++ti;
      --tj;
    }
    while (valid(ti, tj)) {
      left_diag[ti][tj] = 2 * i + 2;
      --ti;
      ++tj;
    }
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      if (i % 2 == 0)
        color[i][j] = j % 2;
      else
        color[i][j] = (j + 1) % 2;
    }
  for (int i = 0; i < n; ++i) {
    int ti = i;
    int tj = n - 1 - i;
    while (valid(ti + 1, tj + 1)) {
      ++ti;
      ++tj;
    }
    while (valid(ti, tj)) {
      right_diag[ti][tj] = 2 * i + 1;
      --ti;
      --tj;
    }
    ti = i;
    tj = n - 1 - i;
    --tj;
    if (!valid(ti, tj)) continue;
    while (valid(ti + 1, tj + 1)) {
      ++ti;
      ++tj;
    }
    while (valid(ti, tj)) {
      right_diag[ti][tj] = 2 * i + 2;
      --ti;
      --tj;
    }
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      left_diag_sum[left_diag[i][j]] += field[i][j];
      right_diag_sum[right_diag[i][j]] += field[i][j];
    }
  pair<int, int> respos[2];
  long long res[2];
  res[0] = -1;
  res[1] = -1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      long long current = left_diag_sum[left_diag[i][j]];
      current += right_diag_sum[right_diag[i][j]];
      current -= field[i][j];
      int c = color[i][j];
      if (res[c] < current) {
        res[c] = current;
        respos[c] = make_pair(i + 1, j + 1);
      }
    }
  cout << res[0] + res[1] << endl;
  cout << respos[0].first << " " << respos[0].second << " " << respos[1].first
       << " " << respos[1].second << endl;
  return 0;
}
