#include <bits/stdc++.h>
using namespace std;
int n, c, x, y;
int ans = 0;
void bs(int l, int r) {
  int m;
  while (l <= r) {
    m = (l + r) / 2;
    long long s = 0;
    for (int i = y; i <= y + m && i <= n; ++i) {
      long long v = (i - y) - m;
      long long le = max(1LL, x + v);
      long long ri = min((long long)n, x - v);
      s += (ri - le) + 1;
    }
    for (int i = y - 1; i >= y - m && i >= 1; --i) {
      long long v = (y - i) - m;
      long long le = max(1LL, x + v);
      long long ri = min((long long)n, x - v);
      s += (ri - le) + 1;
    }
    if (s >= c) {
      ans = m;
      r = m - 1;
    } else
      l = m + 1;
  }
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  bs(0, 100000);
  printf("%d\n", ans);
  return 0;
}

### 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, c, x, y;
int ans = 0;
void bs(int l, int r) {
  int m;
  while (l <= r) {
    m = (l + r) / 2;
    long long s = 0;
    for (int i = y; i <= y + m && i <= n; ++i) {
      long long v = (i - y) - m;
      long long le = max(1LL, x + v);
      long long ri = min((long long)n, x - v);
      s += (ri - le) + 1;
    }
    for (int i = y - 1; i >= y - m && i >= 1; --i) {
      long long v = (y - i) - m;
      long long le = max(1LL, x + v);
      long long ri = min((long long)n, x - v);
      s += (ri - le) + 1;
    }
    if (s >= c) {
      ans = m;
      r = m - 1;
    } else
      l = m + 1;
  }
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  bs(0, 100000);
  printf("%d\n", ans);
  return 0;
}

```