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;
const int N = 2001;
long long a[N][N], b[N][N] = {0}, c[N][N] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int i, j, l, k, x1 = 0, x2 = 0, y1 = 1, y2 = 0, n;
  long long mo = 0, me = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) scanf("%lld", &a[i][j]);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (b[i][j] == 0) {
        l = i;
        k = j;
        while (l + 1 < n && k + 1 < n) b[i][j] += a[l + 1][k + 1], l++, k++;
        l = i;
        k = j;
        while (l - 1 >= 0 && k - 1 >= 0) b[i][j] += a[l - 1][k - 1], l--, k--;
        b[i][j] += a[i][j];
        l = i;
        k = j;
        while (l + 1 < n && k + 1 < n) b[l][k] = b[i][j], l++, k++;
        l = i;
        k = j;
        while (l - 1 >= 0 && k - 1 >= 0) b[l - 1][k - 1] = b[i][j], l--, k--;
      }
      if (c[i][j] == 0) {
        l = i;
        k = j;
        while (l - 1 >= 0 && k + 1 < n) c[i][j] += a[l - 1][k + 1], l--, k++;
        l = i;
        k = j;
        while (l + 1 < n && k - 1 >= 0) c[i][j] += a[l + 1][k - 1], l++, k--;
        c[i][j] += a[i][j];
        l = i;
        k = j;
        while (l - 1 >= 0 && k + 1 < n) c[l - 1][k + 1] = c[i][j], l--, k++;
        l = i;
        k = j;
        while (l + 1 < n && k - 1 >= 0) c[l + 1][k - 1] = c[i][j], l++, k--;
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if ((i + j) % 2 == 1 && mo < b[i][j] + c[i][j] - a[i][j]) {
        mo = b[i][j] + c[i][j] - a[i][j];
        x1 = i;
        y1 = j;
      }
      if ((i + j) % 2 == 0 && me < b[i][j] + c[i][j] - a[i][j]) {
        me = b[i][j] + c[i][j] - a[i][j];
        x2 = i;
        y2 = j;
      }
    }
  }
  printf("%lld\n", mo + me);
  printf("%d %d %d %d\n", x1 + 1, y1 + 1, x2 + 1, y2 + 1);
}
