#include <bits/stdc++.h>
using namespace std;
long long N, X, Y, C;
long long xd[2], yd[2];
long long dlist[4];
long long solve(long long d) {
  long long res = d * d + (d + 1) * (d + 1);
  for (int i = 0; i < 4; i++) {
    if (d >= dlist[i]) {
      res -= (d - dlist[i] + 1) * (d - dlist[i] + 1);
    }
  }
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) {
      if (xd[i] + yd[j] <= d) {
        long long wid = d - xd[i] - yd[j] + 1;
        res += (wid * (wid + 1)) / 2;
      }
    }
  return res;
}
int main() {
  cin >> N >> X >> Y >> C;
  xd[0] = X;
  xd[1] = N - X + 1;
  yd[0] = Y;
  yd[1] = N - Y + 1;
  dlist[0] = xd[0];
  dlist[1] = xd[1];
  dlist[2] = yd[0];
  dlist[3] = yd[1];
  for (long long i = 0; i < 1000000; i++)
    if (solve(i) >= C) {
      cout << i << "\n";
      return 0;
    }
  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>
using namespace std;
long long N, X, Y, C;
long long xd[2], yd[2];
long long dlist[4];
long long solve(long long d) {
  long long res = d * d + (d + 1) * (d + 1);
  for (int i = 0; i < 4; i++) {
    if (d >= dlist[i]) {
      res -= (d - dlist[i] + 1) * (d - dlist[i] + 1);
    }
  }
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) {
      if (xd[i] + yd[j] <= d) {
        long long wid = d - xd[i] - yd[j] + 1;
        res += (wid * (wid + 1)) / 2;
      }
    }
  return res;
}
int main() {
  cin >> N >> X >> Y >> C;
  xd[0] = X;
  xd[1] = N - X + 1;
  yd[0] = Y;
  yd[1] = N - Y + 1;
  dlist[0] = xd[0];
  dlist[1] = xd[1];
  dlist[2] = yd[0];
  dlist[3] = yd[1];
  for (long long i = 0; i < 1000000; i++)
    if (solve(i) >= C) {
      cout << i << "\n";
      return 0;
    }
  return 0;
}

```