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 N = 2005;
long long n, currmax, xcurr, ycurr;
long long arr[N][N], contrib[N][N], vis[N][N];
long long d1[2 * N], d2[2 * N];
vector<pair<long long, pair<long long, long long> > > v;
long long dirx[] = {-1, -1, 1, 1};
long long diry[] = {-1, 1, -1, 1};
void dfs(long long x, long long y) {
  if (x < 1 || y < 1 || x > n || y > n) return;
  if (vis[x][y]) return;
  vis[x][y] = 1;
  if (contrib[x][y] > currmax) {
    currmax = contrib[x][y];
    xcurr = x;
    ycurr = y;
  }
  for (long long i = 0; i < 4; i++) dfs(x + dirx[i], y + diry[i]);
}
void test_case() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cin >> arr[i][j];
      d1[j - i + n] += arr[i][j];
      d2[i + j] += arr[i][j];
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      contrib[i][j] += d1[j - i + n] + d2[i + j] - arr[i][j];
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      if (!vis[i][j]) {
        currmax = -1;
        dfs(i, j);
        v.push_back({currmax, {xcurr, ycurr}});
      }
    }
  }
  sort(v.rbegin(), v.rend());
  cout << v[0].first + v[1].first << endl;
  cout << v[0].second.first << " " << v[0].second.second << " "
       << v[1].second.first << " " << v[1].second.second;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) test_case();
  return 0;
}
