#include <bits/stdc++.h>
using namespace std;
unsigned long long n, c, x, y;
unsigned long long area(unsigned long long time) {
  unsigned long long res = 0ULL;
  time++;
  res = time * time + (time - 1) * (time - 1);
  time--;
  unsigned long long dx, dy;
  dx = dy = 0ULL;
  if (time >= x) dx = time - x;
  if (time >= y) dy = time - y;
  if (dx > 0ULL) res -= dx * dx;
  if (dy > 0ULL) {
    res -= dy * dy;
    dy--;
    unsigned long long tmp = 0ULL;
    if (dy > x) tmp = dy - x;
    if (tmp >= 0ULL) res += tmp * (tmp + 1) >> 1;
    tmp = 0ULL;
    if (dy > n - 1 - x) tmp = dy - (n - 1 - x);
    if (tmp >= 0) res += tmp * (tmp + 1) >> 1;
  }
  dx = dy = 0ULL;
  if (time > n - 1 - x) dx = time - (n - 1 - x);
  if (time > n - 1 - y) dy = time - (n - 1 - y);
  if (dx > 0ULL) res -= dx * dx;
  if (dy > 0ULL) {
    res -= dy * dy;
    dy--;
    unsigned long long tmp = 0ULL;
    if (dy > x) tmp = dy - x;
    if (tmp >= 0) res += tmp * (tmp + 1) >> 1;
    tmp = 0ULL;
    if (dy > n - 1 - x) tmp = dy - (n - 1 - x);
    if (tmp >= 0ULL) res += tmp * (tmp + 1) >> 1;
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  unsigned long long l = 0ULL, r = 2000000001ULL;
  x--, y--;
  while (l < r) {
    unsigned long long mid = l + r >> 1;
    if (area(mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << l << endl;
  return 0;
}

### 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;
unsigned long long n, c, x, y;
unsigned long long area(unsigned long long time) {
  unsigned long long res = 0ULL;
  time++;
  res = time * time + (time - 1) * (time - 1);
  time--;
  unsigned long long dx, dy;
  dx = dy = 0ULL;
  if (time >= x) dx = time - x;
  if (time >= y) dy = time - y;
  if (dx > 0ULL) res -= dx * dx;
  if (dy > 0ULL) {
    res -= dy * dy;
    dy--;
    unsigned long long tmp = 0ULL;
    if (dy > x) tmp = dy - x;
    if (tmp >= 0ULL) res += tmp * (tmp + 1) >> 1;
    tmp = 0ULL;
    if (dy > n - 1 - x) tmp = dy - (n - 1 - x);
    if (tmp >= 0) res += tmp * (tmp + 1) >> 1;
  }
  dx = dy = 0ULL;
  if (time > n - 1 - x) dx = time - (n - 1 - x);
  if (time > n - 1 - y) dy = time - (n - 1 - y);
  if (dx > 0ULL) res -= dx * dx;
  if (dy > 0ULL) {
    res -= dy * dy;
    dy--;
    unsigned long long tmp = 0ULL;
    if (dy > x) tmp = dy - x;
    if (tmp >= 0) res += tmp * (tmp + 1) >> 1;
    tmp = 0ULL;
    if (dy > n - 1 - x) tmp = dy - (n - 1 - x);
    if (tmp >= 0ULL) res += tmp * (tmp + 1) >> 1;
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  unsigned long long l = 0ULL, r = 2000000001ULL;
  x--, y--;
  while (l < r) {
    unsigned long long mid = l + r >> 1;
    if (area(mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << l << endl;
  return 0;
}

```