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;
long long pwr(long long base, long long exp, long long mod = (1000000007LL)) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2) {
      res = (res * base) % mod;
    }
    base = (base * base) % mod;
    exp /= 2;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    gcd(b, a % b);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
const long long int N = 1e5 + 5;
long long int n;
long long int a[2005][2005], d1[N], d2[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long int i = 0; i < n; i++)
    for (long long int j = 0; j < n; j++) {
      cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[i - j + 2000] += a[i][j];
    }
  long long int best1 = -1e12, best2 = -1e12;
  pair<long long int, long long int> res1 = {-1, -1};
  pair<long long int, long long int> res2 = {-1, -1};
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      long long int cur = d1[i + j] + d2[i - j + 2000] - a[i][j];
      if ((i + j) & 1) {
        if (cur > best1) {
          best1 = cur;
          res1 = {i, j};
        }
      } else {
        if (cur > best2) {
          best2 = cur;
          res2 = {i, j};
        }
      }
    }
  }
  cout << best1 + best2 << '\n';
  cout << res1.first + 1 << ' ' << res1.second + 1 << ' ' << res2.first + 1
       << ' ' << res2.second + 1;
  return 0;
}
