#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
inline long long sqr(long long x) { return x <= 0 ? 0 : x * x; }
inline long long s2(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
long long calc(long long t) {
  long long br = 2LL * t * t + 2 * t + 1;
  bool u = t >= y, l = t >= x, d = t >= (n - y + 1), r = t >= (n - x + 1);
  if (u) br -= sqr(t - y + 1);
  if (l) br -= sqr(t - x + 1);
  if (d) br -= sqr(t - (n - y + 1) + 1);
  if (r) br -= sqr(t - (n - x + 1) + 1);
  if (u && l) br += s2(t - y + 1 - x);
  if (l && d) br += s2(t - (n - y + 1) + 1 - x);
  if (d && r) br += s2(t - (n - y + 1) + 1 - (n - x + 1));
  if (r && u) br += s2(t - y + 1 - (n - x + 1));
  if (br < 0) return ~0ULL >> 1;
  return br;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = -1, r = n * 2, m;
  while (l + 1 < r)
    if (calc(m = (l + r) / 2) >= c)
      r = m;
    else
      l = m;
  cout << r << endl;
  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;
long long n, x, y, c;
inline long long sqr(long long x) { return x <= 0 ? 0 : x * x; }
inline long long s2(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
long long calc(long long t) {
  long long br = 2LL * t * t + 2 * t + 1;
  bool u = t >= y, l = t >= x, d = t >= (n - y + 1), r = t >= (n - x + 1);
  if (u) br -= sqr(t - y + 1);
  if (l) br -= sqr(t - x + 1);
  if (d) br -= sqr(t - (n - y + 1) + 1);
  if (r) br -= sqr(t - (n - x + 1) + 1);
  if (u && l) br += s2(t - y + 1 - x);
  if (l && d) br += s2(t - (n - y + 1) + 1 - x);
  if (d && r) br += s2(t - (n - y + 1) + 1 - (n - x + 1));
  if (r && u) br += s2(t - y + 1 - (n - x + 1));
  if (br < 0) return ~0ULL >> 1;
  return br;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = -1, r = n * 2, m;
  while (l + 1 < r)
    if (calc(m = (l + r) / 2) >= c)
      r = m;
    else
      l = m;
  cout << r << endl;
  return 0;
}

```