#include <bits/stdc++.h>
using namespace std;
int long long n;
int long long x, y;
inline int long long u(int long long t) {
  int long long h = t - x + 1;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long r(int long long t) {
  int long long h = t - n + y;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long l(int long long t) {
  int long long h = t - y + 1;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long d(int long long t) {
  int long long h = t - n + x;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long lu(int long long t) {
  int long long h = t + 1 - x - y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
inline int long long ld(int long long t) {
  int long long h = t - n + x - y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
inline int long long ru(int long long t) {
  int h = t - x - n + y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
inline int long long rd(int long long t) {
  int long long h = t - 1 - 2 * n + x + y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
int main() {
  int long long c;
  cin >> n;
  cin >> x >> y >> c;
  int long long t = 0;
  int long long s = 1;
  int long long num = 1;
  while (num < c) {
    t++;
    s += t * 4;
    num = s - l(t) - r(t) - u(t) - d(t) + lu(t) + ld(t) + ru(t) + rd(t);
  }
  cout << t << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int long long n;
int long long x, y;
inline int long long u(int long long t) {
  int long long h = t - x + 1;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long r(int long long t) {
  int long long h = t - n + y;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long l(int long long t) {
  int long long h = t - y + 1;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long d(int long long t) {
  int long long h = t - n + x;
  if (h > 0)
    return h * h;
  else
    return 0;
}
inline int long long lu(int long long t) {
  int long long h = t + 1 - x - y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
inline int long long ld(int long long t) {
  int long long h = t - n + x - y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
inline int long long ru(int long long t) {
  int h = t - x - n + y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
inline int long long rd(int long long t) {
  int long long h = t - 1 - 2 * n + x + y;
  if (h > 0)
    return (1 + h) * h / 2;
  else
    return 0;
}
int main() {
  int long long c;
  cin >> n;
  cin >> x >> y >> c;
  int long long t = 0;
  int long long s = 1;
  int long long num = 1;
  while (num < c) {
    t++;
    s += t * 4;
    num = s - l(t) - r(t) - u(t) - d(t) + lu(t) + ld(t) + ru(t) + rd(t);
  }
  cout << t << endl;
  return 0;
}

```