#include <bits/stdc++.h>
using namespace std;
long long int instante(long long int t, long long int n, long long int x,
                       long long int y) {
  long long int r = 1;
  if (y - t > 1)
    r += t;
  else
    r += y - 1;
  if (y + t > n)
    r += (n - y);
  else
    r += t;
  for (long long int i = 1; i <= t; i++) {
    if (x + i > n) break;
    r++;
    if (y - (t - i) > 1)
      r += t - i;
    else
      r += y - 1;
    if (y + (t - i) > n)
      r += (n - y);
    else
      r += t - i;
  }
  for (long long int i = 1; i <= t; i++) {
    if ((x - i) < 1) break;
    r++;
    if (y - (t - i) > 1)
      r += (t - i);
    else
      r += y - 1;
    if (y + (t - i) > n)
      r += (n - y);
    else
      r += (t - i);
  }
  return r;
}
int main() {
  long long int n, x, y, c, l, r, m, f = -1, s = 1, k;
  cin >> n >> y >> x >> c;
  l = 0;
  r = min(500000LL, 2 * n);
  while (l < r) {
    m = (l + r) / 2;
    if (instante(m, n, x, y) >= c)
      r = m;
    else
      l = m + 1;
  }
  cout << r << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 instante(long long int t, long long int n, long long int x,
                       long long int y) {
  long long int r = 1;
  if (y - t > 1)
    r += t;
  else
    r += y - 1;
  if (y + t > n)
    r += (n - y);
  else
    r += t;
  for (long long int i = 1; i <= t; i++) {
    if (x + i > n) break;
    r++;
    if (y - (t - i) > 1)
      r += t - i;
    else
      r += y - 1;
    if (y + (t - i) > n)
      r += (n - y);
    else
      r += t - i;
  }
  for (long long int i = 1; i <= t; i++) {
    if ((x - i) < 1) break;
    r++;
    if (y - (t - i) > 1)
      r += (t - i);
    else
      r += y - 1;
    if (y + (t - i) > n)
      r += (n - y);
    else
      r += (t - i);
  }
  return r;
}
int main() {
  long long int n, x, y, c, l, r, m, f = -1, s = 1, k;
  cin >> n >> y >> x >> c;
  l = 0;
  r = min(500000LL, 2 * n);
  while (l < r) {
    m = (l + r) / 2;
    if (instante(m, n, x, y) >= c)
      r = m;
    else
      l = m + 1;
  }
  cout << r << endl;
}

```