#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int N = 2005;
int n;
long long a[N][N];
long long dp[N][N][4];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) cin >> a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j][0] = dp[i - 1][j - 1][0] + a[i][j];
  for (int i = n; i >= 1; i--)
    for (int j = 1; j <= n; j++) dp[i][j][1] = dp[i + 1][j + 1][1] + a[i][j];
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= n; i++) dp[i][j][2] = dp[i + 1][j - 1][2] + a[i][j];
  for (int j = n; j >= 1; j--)
    for (int i = 1; i <= n; i++) dp[i][j][3] = dp[i - 1][j + 1][3] + a[i][j];
  long long odd = 0, even = 0;
  int x1, x2, y1, y2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long sum = 0;
      for (int k = 0; k < 4; k++) sum += dp[i][j][k];
      sum -= 3 * a[i][j];
      if ((i + j) % 2) {
        if (odd <= sum) x1 = i, y1 = j, odd = sum;
      } else {
        if (even <= sum) x2 = i, y2 = j, even = sum;
      }
    }
  }
  cout << odd + even << endl;
  cout << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2 << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int N = 2005;
int n;
long long a[N][N];
long long dp[N][N][4];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) cin >> a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j][0] = dp[i - 1][j - 1][0] + a[i][j];
  for (int i = n; i >= 1; i--)
    for (int j = 1; j <= n; j++) dp[i][j][1] = dp[i + 1][j + 1][1] + a[i][j];
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= n; i++) dp[i][j][2] = dp[i + 1][j - 1][2] + a[i][j];
  for (int j = n; j >= 1; j--)
    for (int i = 1; i <= n; i++) dp[i][j][3] = dp[i - 1][j + 1][3] + a[i][j];
  long long odd = 0, even = 0;
  int x1, x2, y1, y2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long sum = 0;
      for (int k = 0; k < 4; k++) sum += dp[i][j][k];
      sum -= 3 * a[i][j];
      if ((i + j) % 2) {
        if (odd <= sum) x1 = i, y1 = j, odd = sum;
      } else {
        if (even <= sum) x2 = i, y2 = j, even = sum;
      }
    }
  }
  cout << odd + even << endl;
  cout << x1 << ' ' << y1 << ' ' << x2 << ' ' << y2 << endl;
}

```