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 modexpo(long long x, long long y, long long m) {
  x = x % m;
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return modexpo(x * x, y / 2, m);
  } else {
    return (x * modexpo(x * x, y / 2, m)) % m;
  }
}
const int mod = 1e9 + 7;
const int maxn = 2e3 + 10;
bool first(int a, int b) { return a > b; }
long long n, m, a[maxn][maxn], d1[2 * maxn], d2[2 * maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      d1[i + j] += a[i][j];
      d2[n + i - j] += a[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      a[i][j] = -a[i][j] + d1[i + j] + d2[n + i - j];
    }
  }
  long long mxe = 0, mxo = 0;
  pair<int, int> x, y;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        mxe = max(mxe, a[i][j]);
        if (mxe == a[i][j]) {
          x.first = i;
          x.second = j;
        }
      } else {
        mxo = max(mxo, a[i][j]);
        if (mxo == a[i][j]) {
          y.first = i;
          y.second = j;
        }
      }
    }
  }
  cout << mxo + mxe << "\n"
       << x.first + 1 << " " << x.second + 1 << " " << y.first + 1 << " "
       << y.second + 1;
  return 0;
}
