#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
long long n, x, y, c;
long long progr(long long first, long long last, long long cnt) {
  long long ans = first + last;
  if (ans % 2 == 0)
    ans /= 2;
  else {
    assert(cnt % 2 == 0);
    cnt /= 2;
  }
  return ans * cnt;
}
long long vuxid1(long long len) {
  if (len <= 0) return 0;
  return progr(1, 1 + 2 * (len - 1), len);
}
long long vuxid2(long long len) {
  if (len <= 0) return 0;
  return progr(1, 1 + 1 * (len - 1), len);
}
long long check(long long mid) {
  long long ans = 1 + progr(0, 4 * mid, mid + 1);
  long long a[4];
  a[0] = x - 1;
  a[1] = n - y;
  a[2] = n - x;
  a[3] = y - 1;
  for (int i = 0; i < 4; ++i) {
    long long tmp = mid - a[i];
    long long pans = vuxid1(mid - a[i]);
    ans -= vuxid1(mid - a[i]);
  }
  for (int i = 0; i < 4; ++i) {
    long long tmp = mid - a[i] - a[(i + 1) % 4] - 1;
    long long pans = vuxid2(mid - a[i] - a[(i + 1) % 4] - 1);
    ans += vuxid2(mid - a[i] - a[(i + 1) % 4] - 1);
  }
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = -1;
  long long r = 2 * n + 1;
  check(11);
  while (l < r - 1) {
    long long mid = (l + r) / 2;
    if (check(mid) >= c)
      r = mid;
    else
      l = mid;
  }
  cout << r << 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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
long long n, x, y, c;
long long progr(long long first, long long last, long long cnt) {
  long long ans = first + last;
  if (ans % 2 == 0)
    ans /= 2;
  else {
    assert(cnt % 2 == 0);
    cnt /= 2;
  }
  return ans * cnt;
}
long long vuxid1(long long len) {
  if (len <= 0) return 0;
  return progr(1, 1 + 2 * (len - 1), len);
}
long long vuxid2(long long len) {
  if (len <= 0) return 0;
  return progr(1, 1 + 1 * (len - 1), len);
}
long long check(long long mid) {
  long long ans = 1 + progr(0, 4 * mid, mid + 1);
  long long a[4];
  a[0] = x - 1;
  a[1] = n - y;
  a[2] = n - x;
  a[3] = y - 1;
  for (int i = 0; i < 4; ++i) {
    long long tmp = mid - a[i];
    long long pans = vuxid1(mid - a[i]);
    ans -= vuxid1(mid - a[i]);
  }
  for (int i = 0; i < 4; ++i) {
    long long tmp = mid - a[i] - a[(i + 1) % 4] - 1;
    long long pans = vuxid2(mid - a[i] - a[(i + 1) % 4] - 1);
    ans += vuxid2(mid - a[i] - a[(i + 1) % 4] - 1);
  }
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = -1;
  long long r = 2 * n + 1;
  check(11);
  while (l < r - 1) {
    long long mid = (l + r) / 2;
    if (check(mid) >= c)
      r = mid;
    else
      l = mid;
  }
  cout << r << endl;
  return 0;
}

```