#include <bits/stdc++.h>
using namespace std;
long long rect(long long h, long long b, long long r) {
  long long aux = (r * (r + 1)) / 2;
  if (r > h) aux -= ((r - h) * (r - h + 1)) / 2;
  if (r > b) aux -= ((r - b) * (r - b + 1)) / 2;
  if (r >= b + h - 1) aux = b * h;
  return aux;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x, y, c, r = 1;
  cin >> n >> x >> y >> c;
  long long a[4];
  a[0] = x, a[1] = y, a[2] = n - x + 1, a[3] = n - y + 1;
  long long p = 1;
  while (p < c) {
    r++;
    long long entran[4];
    p = 0;
    entran[0] = rect(min(a[0], r), min(a[1], r), r);
    entran[1] = rect(min(a[1], r), min(a[2], r), r);
    entran[2] = rect(min(a[2], r), min(a[3], r), r);
    entran[3] = rect(min(a[0], r), min(a[3], r), r);
    for (long long i = (0); i < (long long)(4); i++) {
      p += entran[i];
    }
    for (long long i = (0); i < (long long)(4); i++) p -= min(r, a[i]);
    p += 1;
  }
  cout << r - 1 << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 rect(long long h, long long b, long long r) {
  long long aux = (r * (r + 1)) / 2;
  if (r > h) aux -= ((r - h) * (r - h + 1)) / 2;
  if (r > b) aux -= ((r - b) * (r - b + 1)) / 2;
  if (r >= b + h - 1) aux = b * h;
  return aux;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x, y, c, r = 1;
  cin >> n >> x >> y >> c;
  long long a[4];
  a[0] = x, a[1] = y, a[2] = n - x + 1, a[3] = n - y + 1;
  long long p = 1;
  while (p < c) {
    r++;
    long long entran[4];
    p = 0;
    entran[0] = rect(min(a[0], r), min(a[1], r), r);
    entran[1] = rect(min(a[1], r), min(a[2], r), r);
    entran[2] = rect(min(a[2], r), min(a[3], r), r);
    entran[3] = rect(min(a[0], r), min(a[3], r), r);
    for (long long i = (0); i < (long long)(4); i++) {
      p += entran[i];
    }
    for (long long i = (0); i < (long long)(4); i++) p -= min(r, a[i]);
    p += 1;
  }
  cout << r - 1 << endl;
  return 0;
}

```