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 mp[2030][2030];
long long k1[2030 * 2], k2[2030 * 2], val[2030][2030];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    memset(k1, 0, sizeof(k1));
    memset(k2, 0, sizeof(k2));
    memset(val, 0, sizeof(val));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        scanf("%d", &mp[i][j]);
        k1[i - j + 2030] += mp[i][j];
        k2[i + j] += mp[i][j];
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        val[i][j] = k1[i - j + 2030] + k2[i + j] - mp[i][j];
      }
    }
    int x1, x2, y1, y2;
    long long tem1 = 0, tem2 = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if ((i + j) % 2 == 0) {
          if (val[i][j] >= tem1) {
            x1 = i;
            y1 = j;
            tem1 = val[i][j];
          }
        } else {
          if (val[i][j] >= tem2) {
            x2 = i;
            y2 = j;
            tem2 = val[i][j];
          }
        }
      }
    }
    printf("%I64d\n", tem1 + tem2);
    printf("%d %d %d %d\n", x1, y1, x2, y2);
  }
}
