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 M = 2005;
int a[M][M];
long long L[M << 1];
long long R[M << 1];
int main() {
  int n;
  while (cin >> n) {
    memset(L, 0, sizeof(L));
    memset(R, 0, sizeof(R));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        scanf("%d", &a[i][j]);
        L[i - j + n] += a[i][j];
        R[i + j] += a[i][j];
      }
    }
    int x1, x2, y1, y2;
    long long max1, max2;
    max1 = max2 = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        long long tp = L[i - j + n] + R[i + j] - a[i][j];
        if ((i + j) & 1) {
          if (max1 <= tp) {
            max1 = tp;
            x1 = i;
            y1 = j;
          }
        } else {
          if (max2 <= tp) {
            max2 = tp;
            x2 = i;
            y2 = j;
          }
        }
      }
    }
    cout << max1 + max2 << endl;
    cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
  }
  return 0;
}
