#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long calc(long long sec) {
  long long sum = sec * sec * 2ll + (sec * 2 + 1);
  long long tx = sec - (x - 1);
  long long bx = sec - (n - x);
  long long ty = sec - (y - 1);
  long long by = sec - (n - y);
  if (tx > 0) sum -= tx * tx;
  if (bx > 0) sum -= bx * bx;
  if (ty > 0) {
    sum -= (ty * (ty - 1) + ty);
    long long mt = (ty - 1) - (x - 1), mb = (ty - 1) - (n - x);
    if (mt > 0) sum += (mt * (mt + 1)) / 2;
    if (mb > 0) sum += (mb * (mb + 1)) / 2;
  }
  if (by > 0) {
    sum -= (by * (by - 1) + by);
    long long mt = (by - 1) - (x - 1), mb = (by - 1) - (n - x);
    if (mt > 0) sum += (mt * (mt + 1)) / 2;
    if (mb > 0) sum += (mb * (mb + 1)) / 2;
  }
  return sum;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lb = -1, ub = 2 * n + 1000;
  while (ub - lb > 1) {
    long long mid = (lb + ub) / 2;
    if (calc(mid) >= c)
      ub = mid;
    else
      lb = mid;
  }
  cout << ub << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long n, x, y, c;
long long calc(long long sec) {
  long long sum = sec * sec * 2ll + (sec * 2 + 1);
  long long tx = sec - (x - 1);
  long long bx = sec - (n - x);
  long long ty = sec - (y - 1);
  long long by = sec - (n - y);
  if (tx > 0) sum -= tx * tx;
  if (bx > 0) sum -= bx * bx;
  if (ty > 0) {
    sum -= (ty * (ty - 1) + ty);
    long long mt = (ty - 1) - (x - 1), mb = (ty - 1) - (n - x);
    if (mt > 0) sum += (mt * (mt + 1)) / 2;
    if (mb > 0) sum += (mb * (mb + 1)) / 2;
  }
  if (by > 0) {
    sum -= (by * (by - 1) + by);
    long long mt = (by - 1) - (x - 1), mb = (by - 1) - (n - x);
    if (mt > 0) sum += (mt * (mt + 1)) / 2;
    if (mb > 0) sum += (mb * (mb + 1)) / 2;
  }
  return sum;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lb = -1, ub = 2 * n + 1000;
  while (ub - lb > 1) {
    long long mid = (lb + ub) / 2;
    if (calc(mid) >= c)
      ub = mid;
    else
      lb = mid;
  }
  cout << ub << endl;
  return 0;
}

```