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 n;
int arr[2001][2001];
vector<long long int> main_top, main_bottom;
vector<long long int> pob_top, pob_bottom;
long long int sum(int x, int y) {
  long long int ans = 0;
  if (x <= y) {
    ans += main_top[y - x];
  } else {
    ans += main_bottom[x - y];
  }
  if (x <= n - y - 1) {
    ans += pob_top[x + y];
  } else {
    ans += pob_bottom[x + y + 1 - n];
  }
  return ans - 1ll * arr[x][y];
}
int main() {
  ios::sync_with_stdio(0);
  int i, j, x, y;
  cin >> n;
  main_top.assign(n, 0);
  main_bottom.assign(n, 0);
  pob_top.assign(n, 0);
  pob_bottom.assign(n, 0);
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j) cin >> arr[i][j];
  for (j = 0; j < n; ++j) {
    x = 0;
    y = j;
    while (x < n && y < n) {
      main_top[j] += arr[x][y];
      x++;
      y++;
    }
  }
  for (i = 0; i < n; ++i) {
    x = i;
    y = 0;
    while (x < n && y < n) {
      main_bottom[i] += arr[x][y];
      x++;
      y++;
    }
  }
  for (j = 0; j < n; ++j) {
    x = 0;
    y = j;
    while (x < n && y >= 0) {
      pob_top[j] += arr[x][y];
      x++;
      y--;
    }
  }
  for (i = 0; i < n; ++i) {
    x = i;
    y = n - 1;
    while (x < n && y >= 0) {
      pob_bottom[i] += arr[x][y];
      x++;
      y--;
    }
  }
  int x1, y1, x2, y2;
  long long int MAX1 = -1, MAX2 = -1;
  for (i = 0; i < n; ++i) {
    for (j = 0; j < n; ++j) {
      long long int s = sum(i, j);
      if ((i + j) % 2) {
        if (s > MAX1) {
          MAX1 = s;
          x1 = i + 1;
          y1 = j + 1;
        }
      } else {
        if (s > MAX2) {
          MAX2 = s;
          x2 = i + 1;
          y2 = j + 1;
        }
      }
    }
  }
  cout << MAX1 + MAX2 << endl;
  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  return 0;
}
