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;
long long t[2001][2001];
long long righ[2001][2001];
long long lef[2001][2001];
int dx[] = {1, -1, -1, 1};
int dy[] = {1, 1, -1, -1};
int n;
bool verif(int i, int j) {
  if (i > n - 1 || j > n - 1 || i < 0 || j < 0) {
    return false;
  }
  return true;
}
long long diagl(long long i, long long j) {
  long long a = n - i - 1;
  long long b = j;
  if (a < b) {
    return lef[n - 1][j - a];
  } else
    return lef[i + b][0];
}
long long diagr(long long i, long long j) {
  if (i >= j) {
    return righ[n - 1][j + n - 1 - i];
  } else
    return righ[i + n - 1 - j][n - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  bool bom = true;
  memset(righ, -1, sizeof righ);
  memset(lef, -1, sizeof lef);
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> t[i][j];
    }
  }
  long long a, b, c;
  for (int c = 0; c < 4; c++) {
    for (int s = 0; s < 4; s++) {
      for (int e = 0; e < n; e++) {
        if (s == 0) {
          a = 0;
          b = e;
        }
        if (s == 1) {
          b = 0;
          a = e;
        }
        if (s == 2) {
          b = n - 1;
          a = e;
        }
        if (s == 3) {
          b = e;
          a = n - 1;
        }
        long long q = t[a][b];
        int r = 1;
        while (verif(a + r * dx[c], b + r * dy[c])) {
          q += t[a + r * dx[c]][b + r * dy[c]];
          r++;
        }
        if (c == 0 || c == 2)
          righ[a][b] = max(righ[a][b], q);
        else {
          lef[a][b] = max(lef[a][b], q);
        }
      }
    }
  }
  long long sum;
  long long x1 = 0, y1 = 0, x2 = 0, y2 = 1, black = 0, white = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (j == 0) {
        if (i % 2 == 0) {
          bom = true;
        } else {
          bom = false;
        }
      }
      sum = diagl(i, j) + diagr(i, j) - t[i][j];
      if (bom == true) {
        bom = false;
        if (white < sum) {
          white = sum;
          x1 = i;
          y1 = j;
        }
      } else {
        bom = true;
        if (black < sum) {
          black = sum;
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  cout << black + white << endl;
  cout << x1 + 1 << " " << y1 + 1 << " " << x2 + 1 << " " << y2 + 1;
}
