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 price[4010][4010];
long long ans1[5010], ans2[5010];
long long ans[4010][4010];
long long s;
long long m1, m2;
int main() {
  int n, i, j, x1, y1, x2, y2;
  memset(price, 0, sizeof(price));
  memset(ans, 0, sizeof(ans));
  memset(ans1, 0, sizeof(ans1));
  memset(ans2, 0, sizeof(ans2));
  cin >> n;
  m1 = m2 = -1;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      scanf("%I64d", &price[i][j]);
    }
  }
  for (i = 2; i <= n * 2; i++) {
    s = 0;
    for (j = 1; j <= n; j++) {
      if (i - j < 1 || i - j > n) continue;
      s += price[i - j][j];
    }
    ans1[i - 1] = s;
  }
  for (i = 1 - n; i <= n - 1; i++) {
    s = 0;
    for (j = 1; j <= n; j++) {
      if (j - i < 1 || j - i > n) continue;
      s += price[j][j - i];
    }
    ans2[i + n] = s;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      ans[i][j] = ans1[i + j - 1];
      ans[i][j] -= price[i][j];
      ans[i][j] += ans2[i - j + n];
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        if (m1 < ans[i][j]) {
          m1 = ans[i][j];
          x1 = i;
          y1 = j;
        }
      } else {
        if (m2 < ans[i][j]) {
          m2 = ans[i][j];
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  m1 = m1 + m2;
  cout << m1 << endl;
  cout << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2 << endl;
  return 0;
}
