#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
long long n, x, y, c;
long long getloc(long long a, long long b, long long d) {
  if (a > b) swap(a, b);
  if (a >= d && b >= d) return d * (d + 1) / 2;
  if (a < d && b >= d) return (d - a) * a + a * (a + 1) / 2;
  if (d >= (a + b - 1)) return a * b;
  d = (a + b - 1 - d);
  return a * b - d * (d + 1) / 2;
}
long long get(long long d) {
  return getloc(x - 1, y - 1, max(0ll, d - 1)) +
         getloc(n - x, n - y, max(0ll, d - 1)) +
         getloc(x - 1, n - y, max(0ll, d - 1)) +
         getloc(n - x, y - 1, max(0ll, d - 1)) + min(x - 1, d) + min(y - 1, d) +
         min(n - x, d) + min(n - y, d) + 1;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lb = 0;
  long long rb = 2 * n;
  while (lb < rb) {
    long long mid = (lb + rb) / 2;
    if (get(mid) < c)
      lb = mid + 1;
    else
      rb = mid;
  }
  cout << lb << endl;
  return 0;
}

### 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 double pi = acos(-1.0);
long long n, x, y, c;
long long getloc(long long a, long long b, long long d) {
  if (a > b) swap(a, b);
  if (a >= d && b >= d) return d * (d + 1) / 2;
  if (a < d && b >= d) return (d - a) * a + a * (a + 1) / 2;
  if (d >= (a + b - 1)) return a * b;
  d = (a + b - 1 - d);
  return a * b - d * (d + 1) / 2;
}
long long get(long long d) {
  return getloc(x - 1, y - 1, max(0ll, d - 1)) +
         getloc(n - x, n - y, max(0ll, d - 1)) +
         getloc(x - 1, n - y, max(0ll, d - 1)) +
         getloc(n - x, y - 1, max(0ll, d - 1)) + min(x - 1, d) + min(y - 1, d) +
         min(n - x, d) + min(n - y, d) + 1;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lb = 0;
  long long rb = 2 * n;
  while (lb < rb) {
    long long mid = (lb + rb) / 2;
    if (get(mid) < c)
      lb = mid + 1;
    else
      rb = mid;
  }
  cout << lb << endl;
  return 0;
}

```