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;
struct arr {
  long long s;
  int x, y;
} sum[2];
long long ans, L[2 * 2005], R[2 * 2005], f[2005][2005];
int a[2005][2005], n, i, j;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
      L[i + j - 1] += (long long)a[i][j];
      R[i - j + n] += (long long)a[i][j];
    }
  sum[0] = (arr){0, 1, 1};
  sum[1] = (arr){0, 1, 2};
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) f[i][j] = L[i + j - 1] + R[i - j + n] - a[i][j];
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (f[i][j] > sum[(i + j) & 1].s) {
        sum[(i + j) & 1].s = f[i][j];
        sum[(i + j) & 1].x = i;
        sum[(i + j) & 1].y = j;
      }
  printf("%I64d\n", sum[0].s + sum[1].s);
  printf("%d %d %d %d\n", sum[0].x, sum[0].y, sum[1].x, sum[1].y);
  return 0;
}
