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 int N = 2005, MOD = 1e9 + 7, MAX = 1e5;
int n, m;
int a[N][N];
long long l[N][N], r[N][N];
long long ehseb(int x, int y, int mark) {
  long long ret = 0;
  if (!mark)
    while (y <= n + 1 && x < n) ret += a[x++][y++];
  else
    while (x < n && y >= 0) ret += a[x++][y--];
  return ret;
}
void pree() {
  for (int i = 0; i <= n + 1; i++)
    r[0][i] = ehseb(0, i, 0), l[0][i] = ehseb(0, i, 1);
  for (int i = 0; i < n; i++) {
    r[i][0] = ehseb(i, 0, 0);
    l[i][0] = ehseb(i, 0, 1);
    r[i][n + 1] = ehseb(i, n + 1, 0);
    l[i][n + 1] = ehseb(i, n + 1, 1);
  }
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= n; j++) {
      r[i][j] = r[i - 1][j - 1];
      l[i][j] = l[i - 1][j + 1];
    }
  }
}
priority_queue<pair<long long, pair<int, int> > > q;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= n; j++) scanf("%d", &a[i][j]);
  pree();
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= n; j++) {
      q.push({l[i][j] + r[i][j] - a[i][j], {i + 1, j}});
    }
  }
  pair<long long, pair<int, int> > best1, best2;
  best1 = q.top();
  q.pop();
  bool wa = (best1.second.first + best1.second.second) % 2;
  while (!q.empty()) {
    if ((q.top().second.first + q.top().second.second) % 2 != wa) {
      best2 = q.top();
      break;
    }
    q.pop();
  }
  printf("%lld\n%d %d %d %d", best1.first + best2.first, best1.second.first,
         best1.second.second, best2.second.first, best2.second.second);
  return 0;
}
