#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return "\"" + s + "\""; }
template <typename T>
string to_string(vector<T>);
template <typename U, typename V>
string to_string(pair<U, V> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename T>
string to_string(vector<T> v) {
  string res = "[";
  for (int i = 0; i < (int)v.size() - 1; i++) res += to_string(v[i]) + ", ";
  if (!v.empty()) res += to_string(v.back());
  res += "]";
  return res;
}
long long int dp1[2 * 2005];
long long int dp2[2 * 2005];
int g[2005][2005];
int n;
long long int s1 = -1;
long long int s2 = -1;
int x, y, xz, yz;
int main() {
  scanf("%d", &n);
  memset(dp1, 0, sizeof dp1);
  memset(dp2, 0, sizeof dp2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      scanf("%d", &g[i][j]);
      dp1[i - j + n] += g[i][j];
      dp2[i + j] += g[i][j];
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      long long int f = dp1[i - j + n] + dp2[i + j] - g[i][j];
      if ((i + j) & 1) {
        if (s1 < f) {
          s1 = f;
          x = i;
          y = j;
        }
      } else {
        if (s2 < f) {
          s2 = f;
          xz = i;
          yz = j;
        }
      }
    }
  printf("%lld\n", s1 + s2);
  printf("%d %d %d %d\n", x, y, xz, yz);
}

### Prompt

Please formulate 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>
using namespace std;
string to_string(string s) { return "\"" + s + "\""; }
template <typename T>
string to_string(vector<T>);
template <typename U, typename V>
string to_string(pair<U, V> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename T>
string to_string(vector<T> v) {
  string res = "[";
  for (int i = 0; i < (int)v.size() - 1; i++) res += to_string(v[i]) + ", ";
  if (!v.empty()) res += to_string(v.back());
  res += "]";
  return res;
}
long long int dp1[2 * 2005];
long long int dp2[2 * 2005];
int g[2005][2005];
int n;
long long int s1 = -1;
long long int s2 = -1;
int x, y, xz, yz;
int main() {
  scanf("%d", &n);
  memset(dp1, 0, sizeof dp1);
  memset(dp2, 0, sizeof dp2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      scanf("%d", &g[i][j]);
      dp1[i - j + n] += g[i][j];
      dp2[i + j] += g[i][j];
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      long long int f = dp1[i - j + n] + dp2[i + j] - g[i][j];
      if ((i + j) & 1) {
        if (s1 < f) {
          s1 = f;
          x = i;
          y = j;
        }
      } else {
        if (s2 < f) {
          s2 = f;
          xz = i;
          yz = j;
        }
      }
    }
  printf("%lld\n", s1 + s2);
  printf("%d %d %d %d\n", x, y, xz, yz);
}

```