#include <bits/stdc++.h>
using namespace std;
bool flag(long long n, long long x, long long y, long long c, long long t) {
  long long square = 0, s = t, k;
  square += s * s;
  square += (s + 1) * (s + 1);
  if (x <= t) {
    k = t - x + 1;
    square -= k * k;
  }
  if (x > n - t) {
    k = x + t - n;
    square -= k * k;
  }
  if (y <= t) {
    k = t - y + 1;
    square -= k * k;
  }
  if (y > n - t) {
    k = y + t - n;
    square -= k * k;
  }
  if (t >= x + y) {
    k = t - x - y + 1;
    square += k * (k + 1) / 2;
  }
  if (t - (n - x + 2 + n - y) >= 0) {
    k = t - (n - x + 2 + n - y) + 1;
    square += k * (k + 1) / 2;
  }
  if (t >= x - y + n + 1) {
    k = t - (x + n + 1 - y) + 1;
    square += k * (k + 1) / 2;
  }
  if (t >= n - x + y + 1) {
    k = t - (n - x + y + 1) + 1;
    square += k * (k + 1) / 2;
  }
  return (square >= c);
}
long long run(long long a, long long b, long long n, long long x, long long y,
              long long c) {
  if (a == b) return a;
  long long m = (a + b) / 2;
  if (flag(n, x, y, c, m))
    return run(a, m, n, x, y, c);
  else
    return run(m + 1, b, n, x, y, c);
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  flag(n, x, y, c, 5e8);
  cout << run(0, 1e9, n, x, y, c);
  return 0;
}

### Prompt

Please formulate a cpp solution 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;
bool flag(long long n, long long x, long long y, long long c, long long t) {
  long long square = 0, s = t, k;
  square += s * s;
  square += (s + 1) * (s + 1);
  if (x <= t) {
    k = t - x + 1;
    square -= k * k;
  }
  if (x > n - t) {
    k = x + t - n;
    square -= k * k;
  }
  if (y <= t) {
    k = t - y + 1;
    square -= k * k;
  }
  if (y > n - t) {
    k = y + t - n;
    square -= k * k;
  }
  if (t >= x + y) {
    k = t - x - y + 1;
    square += k * (k + 1) / 2;
  }
  if (t - (n - x + 2 + n - y) >= 0) {
    k = t - (n - x + 2 + n - y) + 1;
    square += k * (k + 1) / 2;
  }
  if (t >= x - y + n + 1) {
    k = t - (x + n + 1 - y) + 1;
    square += k * (k + 1) / 2;
  }
  if (t >= n - x + y + 1) {
    k = t - (n - x + y + 1) + 1;
    square += k * (k + 1) / 2;
  }
  return (square >= c);
}
long long run(long long a, long long b, long long n, long long x, long long y,
              long long c) {
  if (a == b) return a;
  long long m = (a + b) / 2;
  if (flag(n, x, y, c, m))
    return run(a, m, n, x, y, c);
  else
    return run(m + 1, b, n, x, y, c);
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  flag(n, x, y, c, 5e8);
  cout << run(0, 1e9, n, x, y, c);
  return 0;
}

```