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>
long long a[2005][2005];
long long b[2005][2005];
long long c[2005][2005];
long long d[2005][2005];
int main() {
  int n, x, y, x1 = 1, y1 = 1, x2 = 1, y2 = 2;
  long long sum;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) scanf("%I64d", &a[i][j]);
  for (int j = 1; j <= n; j++) {
    x = 1;
    y = j;
    sum = 0;
    while (x <= n && y <= n) {
      sum += a[x][y];
      x++;
      y++;
    }
    x = 1;
    y = j;
    while (x <= n && y <= n) {
      b[x][y] = sum;
      x++;
      y++;
    }
  }
  for (int i = 1; i <= n; i++) {
    x = i;
    y = 1;
    sum = 0;
    while (x <= n && y <= n) {
      sum += a[x][y];
      x++;
      y++;
    }
    x = i;
    y = 1;
    while (x <= n && y <= n) {
      b[x][y] = sum;
      x++;
      y++;
    }
  }
  for (int j = 1; j <= n; j++) {
    x = 1;
    y = j;
    sum = 0;
    while (x <= n && y > 0) {
      sum += a[x][y];
      x++;
      y--;
    }
    x = 1;
    y = j;
    while (x <= n && y > 0) {
      c[x][y] = sum;
      x++;
      y--;
    }
  }
  for (int i = 1; i <= n; i++) {
    x = i;
    y = n;
    sum = 0;
    while (x <= n && y > 0) {
      sum += a[x][y];
      x++;
      y--;
    }
    x = i;
    y = n;
    while (x <= n && y > 0) {
      c[x][y] = sum;
      x++;
      y--;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      d[i][j] = b[i][j] + c[i][j] - a[i][j];
      if ((i + j) % 2 == 0 && d[i][j] > d[x1][y1]) {
        x1 = i;
        y1 = j;
      }
      if ((i + j) % 2 == 1 && d[i][j] > d[x2][y2]) {
        x2 = i;
        y2 = j;
      }
    }
  }
  printf("%I64d\n", d[x1][y1] + d[x2][y2]);
  printf("%d %d %d %d", x1, y1, x2, y2);
  return 0;
}
