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 pnt[2010][2010], lft[2010][2010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (int)0; i < (int)n; i++)
    for (int j = (int)0; j < (int)n; j++) scanf("%I64d", &pnt[i][j]);
  for (int i = (int)0; i < (int)n; i++) lft[i][0] = pnt[i][0];
  for (int j = (int)0; j < (int)n; j++) lft[0][j] = pnt[0][j];
  for (int i = (int)1; i < (int)n; i++) {
    for (int j = (int)1; j < (int)n; j++)
      lft[i][j] = pnt[i][j] + lft[i - 1][j - 1];
  }
  for (int i = (int)n - 2; i >= (int)0; i--) {
    for (int j = (int)n - 2; j >= (int)0; j--) lft[i][j] = lft[i + 1][j + 1];
  }
  for (int i = (int)0; i < (int)n; i++) {
    for (int j = (int)0; j < (int)n; j++) {
      lft[i][j] -= pnt[i][j];
    }
  }
  for (int i = (int)1; i < (int)n; i++) {
    for (int j = (int)n - 2; j >= (int)0; j--) {
      pnt[i][j] = pnt[i][j] + pnt[i - 1][j + 1];
    }
  }
  for (int i = (int)n - 2; i >= (int)0; i--) {
    for (int j = (int)1; j < (int)n; j++) {
      pnt[i][j] = pnt[i + 1][j - 1];
    }
  }
  for (int i = (int)0; i < (int)n; i++) {
    for (int j = (int)0; j < (int)n; j++) {
      pnt[i][j] += lft[i][j];
    }
  }
  int x1 = 0, x2 = 0, y1 = 0, y2 = 1;
  for (int i = 0; i < n; i += 2) {
    for (int j = (int)0; j < (int)n; j++) {
      if (j & 1) {
        if (pnt[i][j] > pnt[x1][y1]) {
          x1 = i;
          y1 = j;
        }
      } else {
        if (pnt[i][j] > pnt[x2][y2]) {
          x2 = i;
          y2 = j;
        }
      }
    }
  }
  for (int i = 1; i < n; i += 2) {
    for (int j = (int)0; j < (int)n; j++) {
      if (j & 1) {
        if (pnt[i][j] > pnt[x2][y2]) {
          x2 = i;
          y2 = j;
        }
      } else {
        if (pnt[i][j] > pnt[x1][y1]) {
          x1 = i;
          y1 = j;
        }
      }
    }
  }
  long long int ans = pnt[x1][y1] + pnt[x2][y2];
  printf("%I64d\n", ans);
  printf("%d %d %d %d\n", x1 + 1, y1 + 1, x2 + 1, y2 + 1);
  return 0;
}
