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 n;
int arr[2002][2002];
long long templ[2002][2002];
long long tempr[2002][2002];
long long left(int x, int y) {
  int i, j, xt = x, yt = y;
  long long sum = 0;
  while (x < n && y >= 0) {
    sum += arr[x][y];
    x++;
    y--;
  }
  return sum;
}
long long right(int x, int y) {
  int i, j, xt = x, yt = y;
  long long sum = 0;
  while (x < n && y < n) {
    sum += arr[x][y];
    x++;
    y++;
  }
  return sum;
}
int main() {
  int i, j, lefts, rights;
  long long sum = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      scanf("%d", &arr[i][j]);
      if (i == j) sum += arr[i][j];
    }
  }
  tempr[0][0] = sum;
  tempr[n - 1][n - 1] = sum;
  for (j = 1; j < n; j++) tempr[0][j] = right(0, j);
  for (i = 1; i < n; i++) tempr[i][0] = right(i, 0);
  for (i = 1; i < n; i++) {
    for (j = 1; j < n; j++) tempr[i][j] = tempr[i - 1][j - 1];
  }
  for (j = 0; j < n; j++) templ[0][j] = left(0, j);
  for (i = 1; i < n; i++) templ[i][n - 1] = left(i, n - 1);
  for (i = 1; i < n; i++) {
    for (j = 0; j < n - 1; j++) templ[i][j] = templ[i - 1][j + 1];
  }
  long long maxi1 = 0, maxi2 = 0;
  bool st = 0;
  pair<int, int> b1, b2;
  for (i = 0; i < n; i++) {
    for (j = st; j < n; j += 2) {
      if (templ[i][j] + tempr[i][j] - arr[i][j] >= maxi1) {
        maxi1 = templ[i][j] + tempr[i][j] - arr[i][j];
        b1.first = i + 1;
        b1.second = j + 1;
      }
    }
    st = 1 - st;
  }
  st = 1;
  for (i = 0; i < n; i++) {
    for (j = st; j < n; j += 2) {
      if (templ[i][j] + tempr[i][j] - arr[i][j] >= maxi2) {
        maxi2 = templ[i][j] + tempr[i][j] - arr[i][j];
        b2.first = i + 1;
        b2.second = j + 1;
      }
    }
    st = 1 - st;
  }
  cout << maxi1 + maxi2 << "\n";
  printf("%d %d %d %d", b1.first, b1.second, b2.first, b2.second);
  return 0;
}
