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 long long mod = 1000000007;
int n;
vector<vector<long long>> num;
long long calc(int x, int y, int dx, int dy) {
  x += dx;
  y += dy;
  long long ret = 0;
  while (x >= 1 && x <= n && y >= 1 && y <= n) {
    ret += num[x][y];
    x += dx;
    y += dy;
  }
  return ret;
}
void solve(int ncase) {
  cin >> n;
  num.assign(n + 1, vector<long long>(n + 1));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> num[i][j];
    }
  }
  long long maxsum[2] = {-1, -1};
  pair<int, int> maxp[2];
  map<pair<int, int>, long long> d1, d2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int i1 = i - min(i, j), j1 = j - min(i, j);
      if (d1.find(pair<int, int>(i1, j1)) == d1.end()) {
        d1[pair<int, int>(i1, j1)] = calc(i1, j1, 1, 1);
      }
      int d = min(i - 0, n + 1 - j);
      int i2 = i - d, j2 = j + d;
      if (d2.find(pair<int, int>(i2, j2)) == d2.end()) {
        d2[pair<int, int>(i2, j2)] = calc(i2, j2, 1, -1);
      }
      long long tmp =
          d1[pair<int, int>(i1, j1)] + d2[pair<int, int>(i2, j2)] - num[i][j];
      if (tmp > maxsum[(i + j) % 2]) {
        maxsum[(i + j) % 2] = tmp;
        maxp[(i + j) % 2] = pair<int, int>(i, j);
      }
    }
  }
  cout << maxsum[0] + maxsum[1] << endl;
  cout << maxp[0].first << " " << maxp[0].second << " " << maxp[1].first << " "
       << maxp[1].second << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int T = 1;
  int ncase = 0;
  while (T--) {
    solve(++ncase);
  }
}
