#include <bits/stdc++.h>
using namespace std;
const long long P = 1000000007LL;
long long N, x, y, c;
long long f(long long k) {
  long long ret = 2 * k * k + 2 * k + 1, a, b, d;
  d = ((x - N) < 0 ? -(x - N) : x - N);
  if (d < k) ret -= ((k - d) * (k - d));
  d = ((x - 1) < 0 ? -(x - 1) : x - 1);
  if (d < k) ret -= ((k - d) * (k - d));
  d = ((y - N) < 0 ? -(y - N) : y - N);
  if (d < k) ret -= ((k - d) * (k - d));
  d = ((y - 1) < 0 ? -(y - 1) : y - 1);
  if (d < k) ret -= ((k - d) * (k - d));
  a = ((x - (N + 1)) < 0 ? -(x - (N + 1)) : x - (N + 1)),
  b = ((y - (N + 1)) < 0 ? -(y - (N + 1)) : y - (N + 1));
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  a = ((x - (N + 1)) < 0 ? -(x - (N + 1)) : x - (N + 1)),
  b = ((y - 0) < 0 ? -(y - 0) : y - 0);
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  a = ((x - 0) < 0 ? -(x - 0) : x - 0),
  b = ((y - (N + 1)) < 0 ? -(y - (N + 1)) : y - (N + 1));
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  a = ((x - 0) < 0 ? -(x - 0) : x - 0), b = ((y - 0) < 0 ? -(y - 0) : y - 0);
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> x >> y >> c;
  long long lo = 0, hi = 2 * N, mit;
  for (; lo < hi;) {
    mit = (lo + hi) >> 1;
    if (c > f(mit))
      lo = mit + 1;
    else
      hi = mit;
  }
  cout << lo << 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;
const long long P = 1000000007LL;
long long N, x, y, c;
long long f(long long k) {
  long long ret = 2 * k * k + 2 * k + 1, a, b, d;
  d = ((x - N) < 0 ? -(x - N) : x - N);
  if (d < k) ret -= ((k - d) * (k - d));
  d = ((x - 1) < 0 ? -(x - 1) : x - 1);
  if (d < k) ret -= ((k - d) * (k - d));
  d = ((y - N) < 0 ? -(y - N) : y - N);
  if (d < k) ret -= ((k - d) * (k - d));
  d = ((y - 1) < 0 ? -(y - 1) : y - 1);
  if (d < k) ret -= ((k - d) * (k - d));
  a = ((x - (N + 1)) < 0 ? -(x - (N + 1)) : x - (N + 1)),
  b = ((y - (N + 1)) < 0 ? -(y - (N + 1)) : y - (N + 1));
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  a = ((x - (N + 1)) < 0 ? -(x - (N + 1)) : x - (N + 1)),
  b = ((y - 0) < 0 ? -(y - 0) : y - 0);
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  a = ((x - 0) < 0 ? -(x - 0) : x - 0),
  b = ((y - (N + 1)) < 0 ? -(y - (N + 1)) : y - (N + 1));
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  a = ((x - 0) < 0 ? -(x - 0) : x - 0), b = ((y - 0) < 0 ? -(y - 0) : y - 0);
  if (a + b <= k) ret += (k - a - b + 1) * (k - a - b + 2) / 2;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> x >> y >> c;
  long long lo = 0, hi = 2 * N, mit;
  for (; lo < hi;) {
    mit = (lo + hi) >> 1;
    if (c > f(mit))
      lo = mit + 1;
    else
      hi = mit;
  }
  cout << lo << endl;
  return 0;
}

```