#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long up, down, l, r;
long long subt(long long days, long long l) {
  if (days > l) {
    return (days - l) * (days - l);
  }
  return 0;
}
long long addi(long long days, long long s1, long long s2) {
  long long s = days - s1 - s2 - 1;
  if (s > 0) return (s * (s + 1)) / 2ll;
  return 0;
}
bool can(long long days) {
  long long tot = days * days + (days + 1) * (days + 1);
  tot -= subt(days, l);
  tot -= subt(days, r);
  tot -= subt(days, up);
  tot -= subt(days, down);
  tot += addi(days, r, up);
  tot += addi(days, r, down);
  tot += addi(days, l, up);
  tot += addi(days, l, down);
  return tot >= c;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> x >> y >> c;
  up = x - 1, down = n - x;
  l = y - 1, r = n - y;
  long long start = 0, end = 1e9;
  while (start <= end) {
    if (start == end || start == end - 1) {
      if (can(start)) {
        cout << start << "\n";
      } else {
        cout << end << "\n";
      }
      return 0;
    }
    long long mid = (start + end) / 2;
    if (can(mid)) {
      end = mid;
    } else {
      start = mid + 1;
    }
  }
}

### Prompt

Develop a solution in cpp 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;
long long n, x, y, c;
long long up, down, l, r;
long long subt(long long days, long long l) {
  if (days > l) {
    return (days - l) * (days - l);
  }
  return 0;
}
long long addi(long long days, long long s1, long long s2) {
  long long s = days - s1 - s2 - 1;
  if (s > 0) return (s * (s + 1)) / 2ll;
  return 0;
}
bool can(long long days) {
  long long tot = days * days + (days + 1) * (days + 1);
  tot -= subt(days, l);
  tot -= subt(days, r);
  tot -= subt(days, up);
  tot -= subt(days, down);
  tot += addi(days, r, up);
  tot += addi(days, r, down);
  tot += addi(days, l, up);
  tot += addi(days, l, down);
  return tot >= c;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> x >> y >> c;
  up = x - 1, down = n - x;
  l = y - 1, r = n - y;
  long long start = 0, end = 1e9;
  while (start <= end) {
    if (start == end || start == end - 1) {
      if (can(start)) {
        cout << start << "\n";
      } else {
        cout << end << "\n";
      }
      return 0;
    }
    long long mid = (start + end) / 2;
    if (can(mid)) {
      end = mid;
    } else {
      start = mid + 1;
    }
  }
}

```