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 dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
int n, a[2020][2020];
long long sum[2020][2020][2];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf("%d", &a[i][j]);
  for (int i = 0; i < n; i++) {
    long long tmp = 0;
    for (int j = 0; j < n; j++)
      if (i - j >= 0) {
        tmp += a[j][i - j];
      }
    for (int j = 0; j < n; j++)
      if (i - j >= 0) {
        sum[j][i - j][0] = tmp;
      }
    tmp = 0;
    for (int j = 0; j < n; j++)
      if (i + j < n) {
        tmp += a[j][i + j];
      }
    for (int j = 0; j < n; j++)
      if (i + j < n) {
        sum[j][i + j][1] = tmp;
      }
  }
  for (int i = 1; i < n; i++) {
    long long tmp = 0;
    for (int j = 0; j < n; j++)
      if (i + j < n) {
        tmp += a[i + j][j];
      }
    for (int j = 0; j < n; j++)
      if (i + j < n) {
        sum[i + j][j][1] = tmp;
      }
    tmp = 0;
    for (int j = 0; j < n; j++)
      if (i + j < n && n - 1 - j >= 0) {
        tmp += a[i + j][n - 1 - j];
      }
    for (int j = 0; j < n; j++)
      if (i + j < n && n - 1 - j >= 0) {
        sum[i + j][n - 1 - j][0] = tmp;
      }
  }
  long long even = -1, odd = -1;
  int x0, y0, x1, y1;
  x0 = y0 = x1 = y1 = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      long long tmp = sum[i][j][0] + sum[i][j][1] - a[i][j];
      if ((i + j) % 2) {
        if (tmp > odd) {
          odd = tmp;
          x0 = i, y0 = j;
        }
      } else {
        if (tmp > even) {
          even = tmp;
          x1 = i, y1 = j;
        }
      }
    }
  }
  cout << even + odd << endl;
  printf("%d %d %d %d\n", x0 + 1, y0 + 1, x1 + 1, y1 + 1);
  return 0;
}
