#include <bits/stdc++.h>
using namespace std;
long long x, y, n, c;
long long sum(long long s) { return s * (s + 1) / 2; }
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long gao(long long k) {
  long long ret = 0;
  if (x - 1 + y - 1 <= k)
    ret = x * y;
  else {
    ret = sum(k + 1);
    if (x <= k) ret -= sum(k - x + 1);
    if (y <= k) ret -= sum(k - y + 1);
  }
  if (x - 1 + n - y <= k)
    ret += x * (n - y + 1);
  else {
    ret += sum(k + 1);
    if (x <= k) ret -= sum(k - x + 1);
    if (y + k > n) ret -= sum(y + k - n);
  }
  if (n - x + y - 1 <= k)
    ret += (n - x + 1) * y;
  else {
    ret += sum(k + 1);
    if (x + k > n) ret -= sum(x + k - n);
    if (y <= k) ret -= sum(k - y + 1);
  }
  if (n - x + n - y <= k)
    ret += (n - x + 1) * (n - y + 1);
  else {
    ret += sum(k + 1);
    if (x + k > n) ret -= sum(x + k - n);
    if (y + k > n) ret -= sum(y + k - n);
  }
  ret -= min(k + 1, x);
  ret -= min(k + 1, y);
  ret -= min(k + 1, n - y + 1);
  ret -= min(k + 1, n - x + 1);
  return ret + 1;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lefts = 0;
  long long rights = 10 * n;
  long long ans = 0;
  while (lefts <= rights) {
    long long mid = (lefts + rights) / 2;
    if (gao(mid) >= c) {
      ans = mid;
      rights = mid - 1;
    } else
      lefts = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a cpp solution 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 x, y, n, c;
long long sum(long long s) { return s * (s + 1) / 2; }
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long gao(long long k) {
  long long ret = 0;
  if (x - 1 + y - 1 <= k)
    ret = x * y;
  else {
    ret = sum(k + 1);
    if (x <= k) ret -= sum(k - x + 1);
    if (y <= k) ret -= sum(k - y + 1);
  }
  if (x - 1 + n - y <= k)
    ret += x * (n - y + 1);
  else {
    ret += sum(k + 1);
    if (x <= k) ret -= sum(k - x + 1);
    if (y + k > n) ret -= sum(y + k - n);
  }
  if (n - x + y - 1 <= k)
    ret += (n - x + 1) * y;
  else {
    ret += sum(k + 1);
    if (x + k > n) ret -= sum(x + k - n);
    if (y <= k) ret -= sum(k - y + 1);
  }
  if (n - x + n - y <= k)
    ret += (n - x + 1) * (n - y + 1);
  else {
    ret += sum(k + 1);
    if (x + k > n) ret -= sum(x + k - n);
    if (y + k > n) ret -= sum(y + k - n);
  }
  ret -= min(k + 1, x);
  ret -= min(k + 1, y);
  ret -= min(k + 1, n - y + 1);
  ret -= min(k + 1, n - x + 1);
  return ret + 1;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lefts = 0;
  long long rights = 10 * n;
  long long ans = 0;
  while (lefts <= rights) {
    long long mid = (lefts + rights) / 2;
    if (gao(mid) >= c) {
      ans = mid;
      rights = mid - 1;
    } else
      lefts = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

```