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 a[2005][2005];
long long b[10020] = {0}, d[10020] = {0};
int main() {
  std::ios_base::sync_with_stdio(0);
  int cases, caseno = 0, n, i, j, k, cnt, x, y, x1, y1, M = 5000;
  long long sum, mx1, mx2;
  cin >> n;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) cin >> a[i][j];
  }
  for (i = 1; i <= n; i++) {
    x = 1, y = i;
    sum = 0;
    while (x <= n && y <= n) sum += a[x++][y++];
    b[M + 1 - i] = sum;
    x = i, y = 1;
    sum = 0;
    while (x <= n && y <= n) sum += a[x++][y++];
    b[M + i - 1] = sum;
    x = 1, y = i;
    sum = 0;
    while (x <= n && y > 0) sum += a[x++][y--];
    d[M + 1 + i] = sum;
    x = i, y = n;
    sum = 0;
    while (x <= n && y > 0) sum += a[x++][y--];
    d[M + i + n] = sum;
  }
  mx1 = mx2 = -1;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      sum = b[M + i - j] + d[M + i + j] - a[i][j];
      if (i % 2 == j % 2) {
        if (sum > mx1) {
          mx1 = sum;
          x1 = i;
          y1 = j;
        }
      } else {
        if (sum > mx2) {
          mx2 = sum;
          x = i;
          y = j;
        }
      }
    }
  }
  cout << mx1 + mx2 << '\n' << x << " " << y << " " << x1 << " " << y1 << '\n';
  return 0;
}
