#include <bits/stdc++.h>
using namespace std;
long long N, X, Y, C;
bool check(long long t) {
  long long cnt = 1 + (t + 1) * t * 2;
  cnt -= ((max(t - abs(X - N), 0LL)) * (max(t - abs(X - N), 0LL)));
  cnt -= ((max(t - abs(Y - N), 0LL)) * (max(t - abs(Y - N), 0LL)));
  cnt -= ((max(t - abs(X - 1), 0LL)) * (max(t - abs(X - 1), 0LL)));
  cnt -= ((max(t - abs(Y - 1), 0LL)) * (max(t - abs(Y - 1), 0LL)));
  cnt += ((max(0LL, t - (abs(X) + abs(Y)) + 1)) *
          (max(0LL, t - (abs(X) + abs(Y)) + 1) + 1) / 2);
  cnt += ((max(0LL, t - (abs(X) + abs(Y - N - 1)) + 1)) *
          (max(0LL, t - (abs(X) + abs(Y - N - 1)) + 1) + 1) / 2);
  cnt += ((max(0LL, t - (abs(X - N - 1) + abs(Y)) + 1)) *
          (max(0LL, t - (abs(X - N - 1) + abs(Y)) + 1) + 1) / 2);
  cnt += ((max(0LL, t - (abs(X - N - 1) + abs(Y - N - 1)) + 1)) *
          (max(0LL, t - (abs(X - N - 1) + abs(Y - N - 1)) + 1) + 1) / 2);
  return cnt >= C;
}
int main() {
  cin >> N >> X >> Y >> C;
  long long lo = -1;
  long long hi = 2 * N;
  while (hi - lo > 1) {
    long long mid = (lo + hi) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid;
  }
  cout << hi << endl;
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

Mr. Bender has a digital table of size n × n, each cell can be switched on or off. He wants the field to have at least c switched on squares. When this condition is fulfilled, Mr Bender will be happy.

We'll consider the table rows numbered from top to bottom from 1 to n, and the columns — numbered from left to right from 1 to n. Initially there is exactly one switched on cell with coordinates (x, y) (x is the row number, y is the column number), and all other cells are switched off. Then each second we switch on the cells that are off but have the side-adjacent cells that are on.

For a cell with coordinates (x, y) the side-adjacent cells are cells with coordinates (x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1).

In how many seconds will Mr. Bender get happy?

Input

The first line contains four space-separated integers n, x, y, c (1 ≤ n, c ≤ 109; 1 ≤ x, y ≤ n; c ≤ n2).

Output

In a single line print a single integer — the answer to the problem.

Examples

Input

6 4 3 1


Output

0


Input

9 3 8 10


Output

2

Note

Initially the first test has one painted cell, so the answer is 0. In the second test all events will go as is shown on the figure. <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long N, X, Y, C;
bool check(long long t) {
  long long cnt = 1 + (t + 1) * t * 2;
  cnt -= ((max(t - abs(X - N), 0LL)) * (max(t - abs(X - N), 0LL)));
  cnt -= ((max(t - abs(Y - N), 0LL)) * (max(t - abs(Y - N), 0LL)));
  cnt -= ((max(t - abs(X - 1), 0LL)) * (max(t - abs(X - 1), 0LL)));
  cnt -= ((max(t - abs(Y - 1), 0LL)) * (max(t - abs(Y - 1), 0LL)));
  cnt += ((max(0LL, t - (abs(X) + abs(Y)) + 1)) *
          (max(0LL, t - (abs(X) + abs(Y)) + 1) + 1) / 2);
  cnt += ((max(0LL, t - (abs(X) + abs(Y - N - 1)) + 1)) *
          (max(0LL, t - (abs(X) + abs(Y - N - 1)) + 1) + 1) / 2);
  cnt += ((max(0LL, t - (abs(X - N - 1) + abs(Y)) + 1)) *
          (max(0LL, t - (abs(X - N - 1) + abs(Y)) + 1) + 1) / 2);
  cnt += ((max(0LL, t - (abs(X - N - 1) + abs(Y - N - 1)) + 1)) *
          (max(0LL, t - (abs(X - N - 1) + abs(Y - N - 1)) + 1) + 1) / 2);
  return cnt >= C;
}
int main() {
  cin >> N >> X >> Y >> C;
  long long lo = -1;
  long long hi = 2 * N;
  while (hi - lo > 1) {
    long long mid = (lo + hi) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid;
  }
  cout << hi << endl;
  return 0;
}

```