#include <bits/stdc++.h>
using namespace std;
const long long inf = 1000000000;
long long n, c, x, y;
long long dist(long long x1, long long y1, long long x2, long long y2) {
  return abs(x1 - x2) + abs(y1 - y2);
}
long long cnt(long long t) {
  if (t == 0) return 1;
  long long sum = t * t + (t + 1) * (t + 1);
  if (x - t < 1) sum -= (t - x + 1) * (t - x + 1);
  if (x + t > n) sum -= (x + t - n) * (x + t - n);
  if (y - t < 1) sum -= (t - y + 1) * (t - y + 1);
  if (y + t > n) sum -= (y + t - n) * (y + t - n);
  long long d;
  d = dist(x, y, 1, 1);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  d = dist(x, y, 1, n);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  d = dist(x, y, n, 1);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  d = dist(x, y, n, n);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  return sum;
}
bool ok(long long t) {
  if (cnt(t) >= (long long)c) return true;
  return false;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0;
  long long r = 2 * n;
  long long ans = 2 * n;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (ok(mid)) {
      ans = min(ans, mid);
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
}

### Prompt

Please create a solution in cpp 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;
const long long inf = 1000000000;
long long n, c, x, y;
long long dist(long long x1, long long y1, long long x2, long long y2) {
  return abs(x1 - x2) + abs(y1 - y2);
}
long long cnt(long long t) {
  if (t == 0) return 1;
  long long sum = t * t + (t + 1) * (t + 1);
  if (x - t < 1) sum -= (t - x + 1) * (t - x + 1);
  if (x + t > n) sum -= (x + t - n) * (x + t - n);
  if (y - t < 1) sum -= (t - y + 1) * (t - y + 1);
  if (y + t > n) sum -= (y + t - n) * (y + t - n);
  long long d;
  d = dist(x, y, 1, 1);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  d = dist(x, y, 1, n);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  d = dist(x, y, n, 1);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  d = dist(x, y, n, n);
  if (d < t) sum += (t - d) * (t - d - 1) / 2;
  return sum;
}
bool ok(long long t) {
  if (cnt(t) >= (long long)c) return true;
  return false;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0;
  long long r = 2 * n;
  long long ans = 2 * n;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (ok(mid)) {
      ans = min(ans, mid);
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
}

```