#include <bits/stdc++.h>
using namespace std;
long long countafter(int x, int y, int t, int n) {
  t--;
  int lowest = max(x - t, 1);
  int highest = min(x + t, n);
  long long answer = 0;
  for (int xp = lowest; xp <= highest; xp++) {
    int miny = max(1, y - t + abs(x - xp));
    int maxy = min(n, y + t - abs(x - xp));
    if (miny > maxy) continue;
    answer += maxy - miny + 1;
  }
  return answer;
}
int main() {
  int n, x, y, c;
  cin >> n >> x >> y >> c;
  if (c == 1) {
    cout << 0 << endl;
    return 0;
  }
  long long low = 0;
  long long high = 1000000;
  while (low < high) {
    long long mid = (low + high) / 2;
    if (countafter(x, y, mid, n) >= c) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  cout << low - 1 << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 countafter(int x, int y, int t, int n) {
  t--;
  int lowest = max(x - t, 1);
  int highest = min(x + t, n);
  long long answer = 0;
  for (int xp = lowest; xp <= highest; xp++) {
    int miny = max(1, y - t + abs(x - xp));
    int maxy = min(n, y + t - abs(x - xp));
    if (miny > maxy) continue;
    answer += maxy - miny + 1;
  }
  return answer;
}
int main() {
  int n, x, y, c;
  cin >> n >> x >> y >> c;
  if (c == 1) {
    cout << 0 << endl;
    return 0;
  }
  long long low = 0;
  long long high = 1000000;
  while (low < high) {
    long long mid = (low + high) / 2;
    if (countafter(x, y, mid, n) >= c) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  cout << low - 1 << endl;
}

```