#include <bits/stdc++.h>
using namespace std;
const int oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-7;
long long n, c, x, y;
long long sub(long long val) {
  if (val <= 0) return 0;
  return val * val;
}
long long add(long long val) {
  if (val <= 0) return 0;
  return val * (val + 1) / 2;
}
bool valid(long long mid) {
  long long ret = (mid + 1) * (mid + 1) * 2 - (mid + 1) * 2 + 1;
  ret -= sub(mid - x + 1);
  ret -= sub(mid - (n - x));
  ret -= sub(mid - y + 1);
  ret -= sub(mid - (n - y));
  ret += add(mid - x - y + 1);
  ret += add(mid - x - (n - y));
  ret += add(mid - (n - x) - y);
  ret += add(mid - (n - x) - (n - y) - 1);
  return ret >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, h = 1e9;
  while (l < h) {
    long long mid = l + (h - l) / 2;
    if (valid(mid))
      h = mid;
    else
      l = mid + 1;
  }
  cout << l << 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;
const int oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-7;
long long n, c, x, y;
long long sub(long long val) {
  if (val <= 0) return 0;
  return val * val;
}
long long add(long long val) {
  if (val <= 0) return 0;
  return val * (val + 1) / 2;
}
bool valid(long long mid) {
  long long ret = (mid + 1) * (mid + 1) * 2 - (mid + 1) * 2 + 1;
  ret -= sub(mid - x + 1);
  ret -= sub(mid - (n - x));
  ret -= sub(mid - y + 1);
  ret -= sub(mid - (n - y));
  ret += add(mid - x - y + 1);
  ret += add(mid - x - (n - y));
  ret += add(mid - (n - x) - y);
  ret += add(mid - (n - x) - (n - y) - 1);
  return ret >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, h = 1e9;
  while (l < h) {
    long long mid = l + (h - l) / 2;
    if (valid(mid))
      h = mid;
    else
      l = mid + 1;
  }
  cout << l << endl;
  return 0;
}

```