#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const long long MAX = 2e3 + 5;
const long long MAX2 = 11;
const int MOD = 1000000000 + 7;
const long long INF = 20000;
const int nr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int nc[] = {0, 1, 0, -1, 1, -1, 1, -1};
long long n, a, b, c, x[MAX][MAX], y[2 * MAX], z[2 * MAX], id1[MAX][MAX],
    id2[MAX][MAX], tmp, cnt, sz, p, q;
string s, t;
pair<int, int> a1, a2;
int main() {
  cout << fixed << setprecision(10);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) cin >> x[i][j];
  c = 2 * n - 1;
  a = n, b = 1;
  for (int i = 1; i <= c; i++) {
    p = 0;
    while (a + p <= n && b + p <= n) {
      id1[a + p][b + p] = i;
      y[i] += x[a + p][b + p];
      p++;
    }
    if (a == 1)
      b++;
    else
      a--;
  }
  a = 1, b = 1;
  for (int i = 1; i <= c; i++) {
    p = 0;
    while (a - p > 0 && b + p <= n) {
      id2[a - p][b + p] = i;
      z[i] += x[a - p][b + p];
      p++;
    }
    if (a == n)
      b++;
    else
      a++;
  }
  a = b = 0;
  a2 = {1, 1}, a1 = {1, 2};
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      tmp = y[id1[i][j]] + z[id2[i][j]] - x[i][j];
      if ((i + j) & 1) {
        if (a < tmp) a = tmp, a1 = {i, j};
      } else {
        if (b < tmp) b = tmp, a2 = {i, j};
      }
    }
  cout << a + b << "\n";
  cout << a2.first << " " << a2.second << " " << a1.first << " " << a1.second
       << "\n";
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const long long MAX = 2e3 + 5;
const long long MAX2 = 11;
const int MOD = 1000000000 + 7;
const long long INF = 20000;
const int nr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int nc[] = {0, 1, 0, -1, 1, -1, 1, -1};
long long n, a, b, c, x[MAX][MAX], y[2 * MAX], z[2 * MAX], id1[MAX][MAX],
    id2[MAX][MAX], tmp, cnt, sz, p, q;
string s, t;
pair<int, int> a1, a2;
int main() {
  cout << fixed << setprecision(10);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) cin >> x[i][j];
  c = 2 * n - 1;
  a = n, b = 1;
  for (int i = 1; i <= c; i++) {
    p = 0;
    while (a + p <= n && b + p <= n) {
      id1[a + p][b + p] = i;
      y[i] += x[a + p][b + p];
      p++;
    }
    if (a == 1)
      b++;
    else
      a--;
  }
  a = 1, b = 1;
  for (int i = 1; i <= c; i++) {
    p = 0;
    while (a - p > 0 && b + p <= n) {
      id2[a - p][b + p] = i;
      z[i] += x[a - p][b + p];
      p++;
    }
    if (a == n)
      b++;
    else
      a++;
  }
  a = b = 0;
  a2 = {1, 1}, a1 = {1, 2};
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      tmp = y[id1[i][j]] + z[id2[i][j]] - x[i][j];
      if ((i + j) & 1) {
        if (a < tmp) a = tmp, a1 = {i, j};
      } else {
        if (b < tmp) b = tmp, a2 = {i, j};
      }
    }
  cout << a + b << "\n";
  cout << a2.first << " " << a2.second << " " << a1.first << " " << a1.second
       << "\n";
  return 0;
}

```