#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const long double pi = acos(-1.0);
long long cut(long long x) { return (x < 0) ? 0 : x * x; }
long long cut(long long x, long long y, long long q) {
  long long tmp = q - (x + y) + 1;
  return (tmp < 0) ? 0 : (tmp * (tmp + 1)) / 2;
}
long long check(long long x, long long y, long long q, long long n) {
  return q * q + (q + 1) * (q + 1) - cut(q - x + 1) - cut(q - (n - x)) -
         cut(q - y + 1) - cut(q - (n - y)) + cut(x, y, q) +
         cut(x, n - y + 1, q) + cut(n - x + 1, y, q) +
         cut(n - x + 1, n - y + 1, q);
}
int main() {
  int n, x, y;
  long long c;
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2 * n;
  while (l < r) {
    long long q = (l + r) / 2;
    if (check(x, y, q, n) >= c)
      r = q;
    else
      l = q + 1;
  }
  cout << r << 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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const long double pi = acos(-1.0);
long long cut(long long x) { return (x < 0) ? 0 : x * x; }
long long cut(long long x, long long y, long long q) {
  long long tmp = q - (x + y) + 1;
  return (tmp < 0) ? 0 : (tmp * (tmp + 1)) / 2;
}
long long check(long long x, long long y, long long q, long long n) {
  return q * q + (q + 1) * (q + 1) - cut(q - x + 1) - cut(q - (n - x)) -
         cut(q - y + 1) - cut(q - (n - y)) + cut(x, y, q) +
         cut(x, n - y + 1, q) + cut(n - x + 1, y, q) +
         cut(n - x + 1, n - y + 1, q);
}
int main() {
  int n, x, y;
  long long c;
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2 * n;
  while (l < r) {
    long long q = (l + r) / 2;
    if (check(x, y, q, n) >= c)
      r = q;
    else
      l = q + 1;
  }
  cout << r << endl;
  return 0;
}

```