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;
using vi = vector<int>;
const int maxn = 4004;
long long int n;
long long int a[2001][2001];
long long int pos[maxn], neg[maxn];
long long int ans, x1, yy1, x2, y2;
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) scanf("%lld", &a[i][j]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      pos[2000 + i - j] += a[i][j];
      neg[i + j] += a[i][j];
    }
  }
  long long int posi = 0;
  ans = INT_MIN;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j + posi <= n; j += 2) {
      if (ans < pos[2000 + i - j - posi] + neg[i + j + posi] - a[i][j + posi]) {
        ans = pos[2000 + i - j - posi] + neg[i + j + posi] - a[i][j + posi];
        x1 = i;
        yy1 = j + posi;
      }
    }
    posi = (posi + 1) % 2;
  }
  posi = 1;
  long long int temp_ans = INT_MIN;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j + posi <= n; j += 2) {
      if (temp_ans <
          pos[2000 + i - j - posi] + neg[i + j + posi] - a[i][j + posi]) {
        temp_ans =
            pos[2000 + i - j - posi] + neg[i + j + posi] - a[i][j + posi];
        x2 = i;
        y2 = j + posi;
      }
    }
    posi = (posi + 1) % 2;
  }
  printf("%lld\n", ans + temp_ans);
  printf("%lld %lld %lld %lld\n", x1, yy1, x2, y2);
  return 0;
}
