#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int aabs(int a) {
  if (a < 0) return -a;
  return a;
}
long long n, x, y, c;
long long fff(long long a) { return (a * (a + 1)) / 2; }
long long ff(long long a) { return a * a; }
bool f(long long m) {
  long long cc = m * m * 2 + 2 * m + 1;
  if (x - m < 1) {
    cc -= ff(-(x - m) + 1);
  }
  if (y - m < 1) {
    cc -= ff(-(y - m) + 1);
  }
  if (x + m > n) {
    cc -= ff(-(n - (x + m)));
  }
  if (y + m > n) {
    cc -= ff(-(n - (y + m)));
  }
  if (x + y - m < 1) {
    cc += fff(-(x + y - m) + 1);
  }
  if ((n - x + 1) + (n - y + 1) - m < 1) {
    cc += fff(-((n - x + 1) + (n - y + 1) - m) + 1);
  }
  if (x + (n - y + 1) - m < 1) {
    cc += fff(-(x + (n - y + 1) - m) + 1);
  }
  if ((n - x + 1) + y - m < 1) {
    cc += fff(-((n - x + 1) + y - m) + 1);
  }
  return cc >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 5000000000ll;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    if (f(m)) {
      r = m;
    } else {
      l = m;
    }
  }
  while (!f(l)) {
    l++;
  }
  cout << l;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int aabs(int a) {
  if (a < 0) return -a;
  return a;
}
long long n, x, y, c;
long long fff(long long a) { return (a * (a + 1)) / 2; }
long long ff(long long a) { return a * a; }
bool f(long long m) {
  long long cc = m * m * 2 + 2 * m + 1;
  if (x - m < 1) {
    cc -= ff(-(x - m) + 1);
  }
  if (y - m < 1) {
    cc -= ff(-(y - m) + 1);
  }
  if (x + m > n) {
    cc -= ff(-(n - (x + m)));
  }
  if (y + m > n) {
    cc -= ff(-(n - (y + m)));
  }
  if (x + y - m < 1) {
    cc += fff(-(x + y - m) + 1);
  }
  if ((n - x + 1) + (n - y + 1) - m < 1) {
    cc += fff(-((n - x + 1) + (n - y + 1) - m) + 1);
  }
  if (x + (n - y + 1) - m < 1) {
    cc += fff(-(x + (n - y + 1) - m) + 1);
  }
  if ((n - x + 1) + y - m < 1) {
    cc += fff(-((n - x + 1) + y - m) + 1);
  }
  return cc >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 5000000000ll;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    if (f(m)) {
      r = m;
    } else {
      l = m;
    }
  }
  while (!f(l)) {
    l++;
  }
  cout << l;
  return 0;
}

```