#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:300000000")
const long double eps = 1e-20;
const long double pi = acos(-1.0);
const long long inf = 1000 * 1000 * 1000 * 1000 * 1000 * 1000;
const long long base = 1000 * 1000 * 1000 + 7;
using namespace std;
long long n, x, y, k;
long long f(long long t) {
  long long res, tmp, lenx, leny;
  res = t * t + (t + 1ll) * (t + 1ll);
  if (x <= t) {
    tmp = t - x + 1ll;
    res -= tmp * tmp;
  }
  if (y <= t) {
    tmp = t - y + 1ll;
    res -= tmp * tmp;
  }
  if (n - x + 1ll <= t) {
    tmp = (t - (n - x + 1ll)) + 1ll;
    res -= tmp * tmp;
  }
  if (n - y + 1ll <= t) {
    tmp = (t - (n - y + 1ll)) + 1ll;
    res -= tmp * tmp;
  }
  if (x < t) {
    lenx = t - x;
    lenx = lenx - y + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  if (n - x + 1ll < t) {
    lenx = t - (n - x + 1ll);
    lenx = lenx - y + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  if (n - y + 1ll < t) {
    lenx = t - (n - y + 1ll);
    lenx = lenx - x + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  if (n - y + 1ll < t) {
    lenx = t - (n - y + 1ll);
    lenx = lenx - (n - x + 1ll) + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> k;
  long long l = 0, r = n + n, c;
  while (r - l > 1) {
    c = (r + l) / 2;
    if (f(c) >= k)
      r = c;
    else
      l = c;
  }
  if (f(l) >= k)
    cout << l;
  else
    cout << r;
  return 0;
}

### Prompt

Generate 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:300000000")
const long double eps = 1e-20;
const long double pi = acos(-1.0);
const long long inf = 1000 * 1000 * 1000 * 1000 * 1000 * 1000;
const long long base = 1000 * 1000 * 1000 + 7;
using namespace std;
long long n, x, y, k;
long long f(long long t) {
  long long res, tmp, lenx, leny;
  res = t * t + (t + 1ll) * (t + 1ll);
  if (x <= t) {
    tmp = t - x + 1ll;
    res -= tmp * tmp;
  }
  if (y <= t) {
    tmp = t - y + 1ll;
    res -= tmp * tmp;
  }
  if (n - x + 1ll <= t) {
    tmp = (t - (n - x + 1ll)) + 1ll;
    res -= tmp * tmp;
  }
  if (n - y + 1ll <= t) {
    tmp = (t - (n - y + 1ll)) + 1ll;
    res -= tmp * tmp;
  }
  if (x < t) {
    lenx = t - x;
    lenx = lenx - y + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  if (n - x + 1ll < t) {
    lenx = t - (n - x + 1ll);
    lenx = lenx - y + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  if (n - y + 1ll < t) {
    lenx = t - (n - y + 1ll);
    lenx = lenx - x + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  if (n - y + 1ll < t) {
    lenx = t - (n - y + 1ll);
    lenx = lenx - (n - x + 1ll) + 1ll;
    if (lenx > 0ll) res += ((lenx) * (lenx + 1ll)) / 2;
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> k;
  long long l = 0, r = n + n, c;
  while (r - l > 1) {
    c = (r + l) / 2;
    if (f(c) >= k)
      r = c;
    else
      l = c;
  }
  if (f(l) >= k)
    cout << l;
  else
    cout << r;
  return 0;
}

```