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;
inline long long msb(long long val) { return 63 - __builtin_clzll(val); }
inline int msb(int val) { return 31 - __builtin_clz(val); }
template <class T>
bool chmin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
template <class T>
bool chmax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
const long long mod = 1e9 + 7, N = 2e3 + 5;
long long a[N][N], n, m, k;
void solve(long long test_case) {
  long long i, j;
  cin >> n;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
    }
  }
  unordered_map<long long, long long> d1, d2;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      d1[i - j] += a[i][j];
      d2[i + j] += a[i][j];
    }
  }
  long long ans1 = -1, ans2 = -1;
  long long x1, y1, x2, y2;
  for (i = 0; i < n; i++) {
    for (j = (i & 1); j < n; j += 2) {
      if (chmax(ans1, d1[i - j] + d2[i + j] - a[i][j])) x1 = i, y1 = j;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 1 - (i & 1); j < n; j += 2) {
      if (chmax(ans2, d1[i - j] + d2[i + j] - a[i][j])) x2 = i, y2 = j;
    }
  }
  cout << ans1 + ans2 << '\n';
  cout << x1 + 1 << ' ' << y1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1 << '\n';
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  solve(1);
  return 0;
}
