#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-3;
const int inf = (1 << 31) - 1;
const int hinf = 0x3f3f3f3f;
const int mod = 1000000007;
int n, x, y, c;
bool okay(int mid) {
  long long cnt = 1ll;
  cnt += (long long)min(x - 1, mid);
  cnt += (long long)min(mid, n - x);
  int le = max(1, y - mid), ri = min(n, y + mid);
  if (cnt >= (long long)c) return 1;
  int now = min(mid - 1, x - 1);
  for (int i = y - 1; i >= le; i--) {
    cnt += (long long)now;
    now = min(mid - (y - i + 1), x - 1);
    if (cnt >= (long long)c) return 1;
  }
  now = min(mid - 1, n - x);
  for (int i = y - 1; i >= le; i--) {
    cnt += (long long)now;
    now = min(mid - (y - i + 1), n - x);
    if (cnt >= (long long)c) return 1;
  }
  now = min(mid - 1, x - 1);
  for (int i = y + 1; i <= ri; i++) {
    cnt += (long long)now;
    now = min(x - 1, mid - (i - y + 1));
    if (cnt >= (long long)c) return 1;
  }
  now = min(n - x, mid - 1);
  for (int i = y + 1; i <= ri; i++) {
    cnt += (long long)now;
    now = min(n - x, mid - (i - y + 1));
    if (cnt >= (long long)c) return 1;
  }
  cnt += (long long)(y - le + ri - y);
  if (cnt >= (long long)c) return 1;
  return 0;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = c, mid, ans;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (okay(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const double eps = 1e-3;
const int inf = (1 << 31) - 1;
const int hinf = 0x3f3f3f3f;
const int mod = 1000000007;
int n, x, y, c;
bool okay(int mid) {
  long long cnt = 1ll;
  cnt += (long long)min(x - 1, mid);
  cnt += (long long)min(mid, n - x);
  int le = max(1, y - mid), ri = min(n, y + mid);
  if (cnt >= (long long)c) return 1;
  int now = min(mid - 1, x - 1);
  for (int i = y - 1; i >= le; i--) {
    cnt += (long long)now;
    now = min(mid - (y - i + 1), x - 1);
    if (cnt >= (long long)c) return 1;
  }
  now = min(mid - 1, n - x);
  for (int i = y - 1; i >= le; i--) {
    cnt += (long long)now;
    now = min(mid - (y - i + 1), n - x);
    if (cnt >= (long long)c) return 1;
  }
  now = min(mid - 1, x - 1);
  for (int i = y + 1; i <= ri; i++) {
    cnt += (long long)now;
    now = min(x - 1, mid - (i - y + 1));
    if (cnt >= (long long)c) return 1;
  }
  now = min(n - x, mid - 1);
  for (int i = y + 1; i <= ri; i++) {
    cnt += (long long)now;
    now = min(n - x, mid - (i - y + 1));
    if (cnt >= (long long)c) return 1;
  }
  cnt += (long long)(y - le + ri - y);
  if (cnt >= (long long)c) return 1;
  return 0;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = c, mid, ans;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (okay(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
}

```