#include <bits/stdc++.h>
using namespace std;
long long int x, y, n, c;
long long l, r, u, d;
bool isok(long long mid) {
  long long have = 2 * mid + 2 * mid * mid + 1;
  long long lmax = (mid - l > 0) ? mid - l : 0;
  long long rmax = (mid - r > 0) ? mid - r : 0;
  long long dmax = (mid - d > 0) ? mid - d : 0;
  long long umax = (mid - u > 0) ? mid - u : 0;
  have -= umax * umax + dmax * dmax + lmax * lmax + rmax * rmax;
  if (umax - y > 0) {
    long long ll = umax - y;
    have += (ll + 1) * ll / 2;
  }
  if (dmax - (n - y + 1) > 0) {
    long long ll = dmax - (n - y + 1);
    have += (ll + 1) * ll / 2;
  }
  if (umax - (n - y + 1) > 0) {
    long long ll = umax - (n - y + 1);
    have += (ll + 1) * ll / 2;
  }
  if (dmax - y > 0) {
    long long ll = dmax - y;
    have += (ll + 1) * ll / 2;
  }
  if (have >= c)
    return true;
  else
    return false;
}
int main(int argc, char *argv[]) {
  cin >> n >> x >> y >> c;
  long long ans = 0;
  l = y - 1;
  r = n - y;
  u = x - 1;
  d = n - x;
  long long low = -1;
  long long high = 2 * n;
  while (low < high - 1) {
    long long mid = (high + low) / 2;
    if (isok(mid))
      high = mid;
    else
      low = mid;
  }
  cout << high << 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>
using namespace std;
long long int x, y, n, c;
long long l, r, u, d;
bool isok(long long mid) {
  long long have = 2 * mid + 2 * mid * mid + 1;
  long long lmax = (mid - l > 0) ? mid - l : 0;
  long long rmax = (mid - r > 0) ? mid - r : 0;
  long long dmax = (mid - d > 0) ? mid - d : 0;
  long long umax = (mid - u > 0) ? mid - u : 0;
  have -= umax * umax + dmax * dmax + lmax * lmax + rmax * rmax;
  if (umax - y > 0) {
    long long ll = umax - y;
    have += (ll + 1) * ll / 2;
  }
  if (dmax - (n - y + 1) > 0) {
    long long ll = dmax - (n - y + 1);
    have += (ll + 1) * ll / 2;
  }
  if (umax - (n - y + 1) > 0) {
    long long ll = umax - (n - y + 1);
    have += (ll + 1) * ll / 2;
  }
  if (dmax - y > 0) {
    long long ll = dmax - y;
    have += (ll + 1) * ll / 2;
  }
  if (have >= c)
    return true;
  else
    return false;
}
int main(int argc, char *argv[]) {
  cin >> n >> x >> y >> c;
  long long ans = 0;
  l = y - 1;
  r = n - y;
  u = x - 1;
  d = n - x;
  long long low = -1;
  long long high = 2 * n;
  while (low < high - 1) {
    long long mid = (high + low) / 2;
    if (isok(mid))
      high = mid;
    else
      low = mid;
  }
  cout << high << endl;
  return 0;
}

```