#include <bits/stdc++.h>
using namespace std;
int n, x, y, c;
long long S(long long n) { return n * (n + 1) / 2; }
long long solve(int m, int maxX, int maxY) {
  if (maxX + maxY <= m) {
    return ((long long)maxX + 1) * ((long long)maxY + 1);
  }
  long long totalArea = S(m + 1);
  if (m > maxX) {
    int diff = m - maxX;
    totalArea -= S(diff);
  }
  if (m > maxY) {
    int diff = m - maxY;
    totalArea -= S(diff);
  }
  return totalArea;
}
int main() {
  cin >> n >> x >> y >> c;
  long long res = 0;
  int l = 0, r = 2 * n;
  while (l <= r) {
    int m = (int)((long long)l + r) / 2;
    long long totalArea = solve(m, n - x, n - y) + solve(m, x - 1, n - y) +
                          solve(m, n - x, y - 1) + solve(m, x - 1, y - 1) -
                          min(m + 1, n - x + 1) - min(m + 1, n - y + 1) -
                          min(m + 1, x) - min(m + 1, y) + 1;
    if (totalArea >= c) {
      r = m - 1;
      res = m;
    } else {
      l = m + 1;
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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;
int n, x, y, c;
long long S(long long n) { return n * (n + 1) / 2; }
long long solve(int m, int maxX, int maxY) {
  if (maxX + maxY <= m) {
    return ((long long)maxX + 1) * ((long long)maxY + 1);
  }
  long long totalArea = S(m + 1);
  if (m > maxX) {
    int diff = m - maxX;
    totalArea -= S(diff);
  }
  if (m > maxY) {
    int diff = m - maxY;
    totalArea -= S(diff);
  }
  return totalArea;
}
int main() {
  cin >> n >> x >> y >> c;
  long long res = 0;
  int l = 0, r = 2 * n;
  while (l <= r) {
    int m = (int)((long long)l + r) / 2;
    long long totalArea = solve(m, n - x, n - y) + solve(m, x - 1, n - y) +
                          solve(m, n - x, y - 1) + solve(m, x - 1, y - 1) -
                          min(m + 1, n - x + 1) - min(m + 1, n - y + 1) -
                          min(m + 1, x) - min(m + 1, y) + 1;
    if (totalArea >= c) {
      r = m - 1;
      res = m;
    } else {
      l = m + 1;
    }
  }
  cout << res << endl;
  return 0;
}

```