#include <bits/stdc++.h>
using namespace std;
int n, c, x, y;
long long g(long long a) {
  if (a <= 0) return 0;
  return a * a;
}
long long f(long long a) {
  if (a <= 0) return 0;
  return (a * (a + 1)) / 2;
}
long long T(long long t) { return 2 * t * t + 2 * t + 1; }
long long calc(int t) {
  int a1, a2, b1, b2;
  b1 = max(1, x - t);
  b2 = min(x + t, n);
  a1 = max(1, y - t);
  a2 = min(n, y + t);
  long long cnt = T(t);
  cnt -= g(y + t - a2);
  cnt -= g(a1 - y + t);
  cnt -= g(x + t - b2);
  cnt -= g(b1 - x + t);
  cnt += f(t - (x - b1) - (a2 - y) - 1);
  cnt += f(t - (x - b1) - (y - a1) - 1);
  cnt += f(t - (b2 - x) - (a2 - y) - 1);
  cnt += f(t - (b2 - x) - (y - a1) - 1);
  return cnt;
}
int main(int nargs, char* args[]) {
  cin >> n >> x >> y >> c;
  if (c == 1) {
    puts("0");
    return 0;
  }
  int lo = 0, hi = 2 * n, mid;
  while (hi - lo > 1) {
    mid = (lo + hi) / 2;
    if (calc(mid) < c)
      lo = mid;
    else
      hi = mid;
  }
  cout << hi << endl;
}

### Prompt

Please formulate 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;
int n, c, x, y;
long long g(long long a) {
  if (a <= 0) return 0;
  return a * a;
}
long long f(long long a) {
  if (a <= 0) return 0;
  return (a * (a + 1)) / 2;
}
long long T(long long t) { return 2 * t * t + 2 * t + 1; }
long long calc(int t) {
  int a1, a2, b1, b2;
  b1 = max(1, x - t);
  b2 = min(x + t, n);
  a1 = max(1, y - t);
  a2 = min(n, y + t);
  long long cnt = T(t);
  cnt -= g(y + t - a2);
  cnt -= g(a1 - y + t);
  cnt -= g(x + t - b2);
  cnt -= g(b1 - x + t);
  cnt += f(t - (x - b1) - (a2 - y) - 1);
  cnt += f(t - (x - b1) - (y - a1) - 1);
  cnt += f(t - (b2 - x) - (a2 - y) - 1);
  cnt += f(t - (b2 - x) - (y - a1) - 1);
  return cnt;
}
int main(int nargs, char* args[]) {
  cin >> n >> x >> y >> c;
  if (c == 1) {
    puts("0");
    return 0;
  }
  int lo = 0, hi = 2 * n, mid;
  while (hi - lo > 1) {
    mid = (lo + hi) / 2;
    if (calc(mid) < c)
      lo = mid;
    else
      hi = mid;
  }
  cout << hi << endl;
}

```