#include <bits/stdc++.h>
using namespace std;
string itos(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
long long n, x, y, c;
long long getarea(long long h1, long long h2, long long len) {
  return (h1 + h2) * len / 2;
}
bool chk(int k) {
  long long tt = 1LL * (k + 1) * (k + 1) + 1LL * k * k;
  if (y <= k) {
    long long d = k - y;
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (y + k > n) {
    long long d = k - (n + 1 - y);
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (x <= k) {
    long long d = k - x;
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (x + k > n) {
    long long d = k - (n + 1 - x);
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (x + y <= k) {
    long long d = k - (x + y);
    tt += (d + 1) * (d + 2) / 2;
  }
  if (x + (n + 1 - y) <= k) {
    long long d = k - (x + (n + 1 - y));
    tt += (d + 1) * (d + 2) / 2;
  }
  if ((n + 1 - x) + y <= k) {
    long long d = k - (y + (n + 1 - x));
    tt += (d + 1) * (d + 2) / 2;
  }
  if ((n + 1 - x) + (n + 1 - y) <= k) {
    long long d = k - ((n + 1 - x) + (n + 1 - y));
    tt += (d + 1) * (d + 2) / 2;
  }
  return tt >= c;
}
int bins(int l, int r) {
  int mid = (l + r) / 2;
  if (l > r) return l;
  if (chk(mid))
    return bins(l, mid - 1);
  else
    return bins(mid + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> x >> y >> c;
  cout << bins(0, (int)(1e9)) << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
string itos(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
long long n, x, y, c;
long long getarea(long long h1, long long h2, long long len) {
  return (h1 + h2) * len / 2;
}
bool chk(int k) {
  long long tt = 1LL * (k + 1) * (k + 1) + 1LL * k * k;
  if (y <= k) {
    long long d = k - y;
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (y + k > n) {
    long long d = k - (n + 1 - y);
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (x <= k) {
    long long d = k - x;
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (x + k > n) {
    long long d = k - (n + 1 - x);
    tt -= (2 * (d + 1)) * (d + 1) / 2;
  }
  if (x + y <= k) {
    long long d = k - (x + y);
    tt += (d + 1) * (d + 2) / 2;
  }
  if (x + (n + 1 - y) <= k) {
    long long d = k - (x + (n + 1 - y));
    tt += (d + 1) * (d + 2) / 2;
  }
  if ((n + 1 - x) + y <= k) {
    long long d = k - (y + (n + 1 - x));
    tt += (d + 1) * (d + 2) / 2;
  }
  if ((n + 1 - x) + (n + 1 - y) <= k) {
    long long d = k - ((n + 1 - x) + (n + 1 - y));
    tt += (d + 1) * (d + 2) / 2;
  }
  return tt >= c;
}
int bins(int l, int r) {
  int mid = (l + r) / 2;
  if (l > r) return l;
  if (chk(mid))
    return bins(l, mid - 1);
  else
    return bins(mid + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> x >> y >> c;
  cout << bins(0, (int)(1e9)) << endl;
  return 0;
}

```