#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
inline long long get1(long long t) {
  if (t < 0) return 0;
  return (t + 1) * (t + 1);
}
inline long long get2(long long t) {
  if (t < 0) return 0;
  return (t + 1) * (t + 2) / 2;
}
inline bool check(long long t) {
  long long res = (2 * t + 1) + 2 * t * t;
  res -=
      get1(t - x) + get1(t - y) + get1(t - (n + 1 - x)) + get1(t - (n + 1 - y));
  res += get2(t - x - y) + get2(t - x - (n + 1 - y)) +
         get2(t - (n + 1 - x) - y) + get2(t - (n + 1 - x) - (n + 1 - y));
  if (res >= c) return true;
  return false;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x >> y >> c;
  long long l = -1, r = 2 * n;
  while (l < r - 1) {
    long long mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  cout << r << '\n';
}

### Prompt

Create a solution in CPP for 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;
long long n, x, y, c;
inline long long get1(long long t) {
  if (t < 0) return 0;
  return (t + 1) * (t + 1);
}
inline long long get2(long long t) {
  if (t < 0) return 0;
  return (t + 1) * (t + 2) / 2;
}
inline bool check(long long t) {
  long long res = (2 * t + 1) + 2 * t * t;
  res -=
      get1(t - x) + get1(t - y) + get1(t - (n + 1 - x)) + get1(t - (n + 1 - y));
  res += get2(t - x - y) + get2(t - x - (n + 1 - y)) +
         get2(t - (n + 1 - x) - y) + get2(t - (n + 1 - x) - (n + 1 - y));
  if (res >= c) return true;
  return false;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x >> y >> c;
  long long l = -1, r = 2 * n;
  while (l < r - 1) {
    long long mid = (l + r) >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  cout << r << '\n';
}

```