#include <bits/stdc++.h>
using namespace std;
const long long INF = (1LL << 45LL);
const long long MAXLL = 9223372036854775807LL;
const unsigned long long MAXULL = 18446744073709551615LLU;
const long long MOD = 100000000;
const long double DELTA = 0.000000001L;
long long mpow(long long x, long long n1) {
  long long ret = 1;
  while (n1 > 0) {
    if (n1 % 2 == 1) ret = ret * x % MOD;
    x = x * x % MOD;
    n1 /= 2;
  }
  return ret;
}
long long inverse(long long x) { return mpow(x, MOD - 2); }
long long dx[4] = {0, 0, 1, -1};
long long dy[4] = {1, -1, 0, 0};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  long long n;
  cin >> n;
  long long A[n][n];
  map<long long, long long> mp1, mp2;
  for (long long i = 0; i < (n); ++i) {
    for (long long j = 0; j < (n); ++j) {
      cin >> A[i][j];
      mp1[i + j] = mp1[i + j] + A[i][j];
      mp2[i - j] = mp2[i - j] + A[i][j];
    }
  }
  long long mx1 = 0, mx2 = 0, i1 = 0, i2 = 0, j1 = 0, j2 = 1;
  for (long long i = 0; i < (n); ++i) {
    for (long long j = 0; j < (n); ++j) {
      if ((i + j) & 1) {
        if (mx2 < mp1[i + j] + mp2[i - j] - A[i][j]) {
          i2 = i;
          j2 = j;
          mx2 = mp1[i + j] + mp2[i - j] - A[i][j];
        }
      } else {
        if (mx1 < mp1[i + j] + mp2[i - j] - A[i][j]) {
          i1 = i;
          j1 = j;
          mx1 = mp1[i + j] + mp2[i - j] - A[i][j];
        }
      }
    }
  }
  cout << mx1 + mx2 << endl;
  cout << i1 + 1 << " " << j1 + 1 << " " << i2 + 1 << " " << j2 + 1 << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const long long INF = (1LL << 45LL);
const long long MAXLL = 9223372036854775807LL;
const unsigned long long MAXULL = 18446744073709551615LLU;
const long long MOD = 100000000;
const long double DELTA = 0.000000001L;
long long mpow(long long x, long long n1) {
  long long ret = 1;
  while (n1 > 0) {
    if (n1 % 2 == 1) ret = ret * x % MOD;
    x = x * x % MOD;
    n1 /= 2;
  }
  return ret;
}
long long inverse(long long x) { return mpow(x, MOD - 2); }
long long dx[4] = {0, 0, 1, -1};
long long dy[4] = {1, -1, 0, 0};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  long long n;
  cin >> n;
  long long A[n][n];
  map<long long, long long> mp1, mp2;
  for (long long i = 0; i < (n); ++i) {
    for (long long j = 0; j < (n); ++j) {
      cin >> A[i][j];
      mp1[i + j] = mp1[i + j] + A[i][j];
      mp2[i - j] = mp2[i - j] + A[i][j];
    }
  }
  long long mx1 = 0, mx2 = 0, i1 = 0, i2 = 0, j1 = 0, j2 = 1;
  for (long long i = 0; i < (n); ++i) {
    for (long long j = 0; j < (n); ++j) {
      if ((i + j) & 1) {
        if (mx2 < mp1[i + j] + mp2[i - j] - A[i][j]) {
          i2 = i;
          j2 = j;
          mx2 = mp1[i + j] + mp2[i - j] - A[i][j];
        }
      } else {
        if (mx1 < mp1[i + j] + mp2[i - j] - A[i][j]) {
          i1 = i;
          j1 = j;
          mx1 = mp1[i + j] + mp2[i - j] - A[i][j];
        }
      }
    }
  }
  cout << mx1 + mx2 << endl;
  cout << i1 + 1 << " " << j1 + 1 << " " << i2 + 1 << " " << j2 + 1 << endl;
  return 0;
}

```