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 maxn = 2000 + 10;
long long int sum[4][maxn][maxn], a[maxn][maxn];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) scanf("%d", a[i] + j);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      sum[0][i][j] = sum[0][i - 1][j - 1] + a[i - 1][j - 1];
  for (int i = 1; i <= n; i++)
    for (int j = n; j >= 1; j--)
      sum[1][i][j] = sum[1][i - 1][j + 1] + a[i - 1][j + 1];
  for (int i = n; i >= 1; i--)
    for (int j = 1; j <= n; j++)
      sum[2][i][j] = sum[2][i + 1][j - 1] + a[i + 1][j - 1];
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--)
      sum[3][i][j] = sum[3][i + 1][j + 1] + a[i + 1][j + 1];
  int x = 0, y = 0, xx = 0, yy = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k < 4; k++) sum[0][i][j] += sum[k][i][j];
      sum[0][i][j] += a[i][j];
      if ((i + j) & 1) {
        if (sum[0][i][j] >= sum[0][x][y]) x = i, y = j;
      } else if (sum[0][i][j] >= sum[0][xx][yy])
        xx = i, yy = j;
    }
  printf("%I64d\n%d %d %d %d", sum[0][x][y] + sum[0][xx][yy], x, y, xx, yy);
  return 0;
}
