#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:64000000")
long long res;
long long n, x, y, c;
inline long long abs64(long long x) {
  if (x < 0) return -x;
  return x;
}
inline long long getCount(long long a, long long b, long long t) {
  t -= abs64(a - b);
  if (t < 1) return 0LL;
  return t * t;
}
inline long long getLadder(long long x) {
  if (x < 1) return 0LL;
  return x * (x + 1) / 2;
}
inline long long getCorners(long long x, long long y, long long d,
                            long long t) {
  t -= abs64(d - x);
  if (t <= 0) return 0LL;
  long long res = 0LL;
  res += getLadder(t - (n - y + 1));
  res += getLadder(t - y);
  return res;
}
inline long long count(long long t) {
  long long res = t * t + (t + 1) * (t + 1);
  res -= getCount(y, n, t);
  res -= getCount(y, 1, t);
  res += getCorners(x, y, n, t);
  res += getCorners(x, y, 1, t);
  res -= getCount(x, n, t);
  res -= getCount(x, 1, t);
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0LL, r = 1000000005LL;
  res = r;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    long long cur = count(mid);
    if (cur >= c) {
      res = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
#pragma comment(linker, "/STACK:64000000")
long long res;
long long n, x, y, c;
inline long long abs64(long long x) {
  if (x < 0) return -x;
  return x;
}
inline long long getCount(long long a, long long b, long long t) {
  t -= abs64(a - b);
  if (t < 1) return 0LL;
  return t * t;
}
inline long long getLadder(long long x) {
  if (x < 1) return 0LL;
  return x * (x + 1) / 2;
}
inline long long getCorners(long long x, long long y, long long d,
                            long long t) {
  t -= abs64(d - x);
  if (t <= 0) return 0LL;
  long long res = 0LL;
  res += getLadder(t - (n - y + 1));
  res += getLadder(t - y);
  return res;
}
inline long long count(long long t) {
  long long res = t * t + (t + 1) * (t + 1);
  res -= getCount(y, n, t);
  res -= getCount(y, 1, t);
  res += getCorners(x, y, n, t);
  res += getCorners(x, y, 1, t);
  res -= getCount(x, n, t);
  res -= getCount(x, 1, t);
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0LL, r = 1000000005LL;
  res = r;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    long long cur = count(mid);
    if (cur >= c) {
      res = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << res << endl;
  return 0;
}

```