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 int b[4001], c[4001], a[2000][2000], mmax[2], t;
int main(int argc, char **argv) {
  for (int i = 0; i < 4001; i++) b[i] = c[i] = 0;
  int n, mx[2], my[2], tch;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      scanf("%I64d", &a[i][j]);
      b[i - j + 2000] += a[i][j];
      c[i + j] += a[i][j];
    }
  mmax[0] = mmax[1] = -1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      t = b[i - j + 2000] + c[i + j] - a[i][j];
      tch = (i + j) % 2;
      if (mmax[tch] < t) {
        mmax[tch] = t;
        mx[tch] = i + 1;
        my[tch] = j + 1;
      }
    }
  printf("%I64d\n%d %d %d %d", mmax[1] + mmax[0], mx[0], my[0], mx[1], my[1]);
  return 0;
}
