#include <bits/stdc++.h>
using namespace std;
int n, x, y, c;
bool chk(int);
int main() {
  int ans, l, r, m;
  scanf("%d %d %d %d", &n, &x, &y, &c);
  l = 0;
  r = 45000;
  while (l <= r) {
    m = (l + r) >> 1;
    if (chk(m)) {
      r = m - 1;
      ans = m;
    } else
      l = m + 1;
  }
  printf("%d\n", ans);
  return 0;
}
bool chk(int l) {
  int i, h, t, ans = 0;
  for (i = max(1, x - l); i <= min(n, x + l); ++i) {
    h = min(n, y + (l - ((x - i) < 0 ? -(x - i) : (x - i))));
    t = max(1, y - (l - ((x - i) < 0 ? -(x - i) : (x - i))));
    ans += h - t + 1;
  }
  return ans >= c;
}

### Prompt

Your challenge is to write 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;
bool chk(int);
int main() {
  int ans, l, r, m;
  scanf("%d %d %d %d", &n, &x, &y, &c);
  l = 0;
  r = 45000;
  while (l <= r) {
    m = (l + r) >> 1;
    if (chk(m)) {
      r = m - 1;
      ans = m;
    } else
      l = m + 1;
  }
  printf("%d\n", ans);
  return 0;
}
bool chk(int l) {
  int i, h, t, ans = 0;
  for (i = max(1, x - l); i <= min(n, x + l); ++i) {
    h = min(n, y + (l - ((x - i) < 0 ? -(x - i) : (x - i))));
    t = max(1, y - (l - ((x - i) < 0 ? -(x - i) : (x - i))));
    ans += h - t + 1;
  }
  return ans >= c;
}

```