#include <bits/stdc++.h>
using namespace std;
int arr[5000];
int M[1000010];
int P[1000010];
int ST[1000010];
long long funct(long long mid, long long X, long long Y, long long N) {
  long long e = (mid * mid + (mid + 1) * (mid + 1));
  long long left = mid - (X - 1);
  left = (left < 0 ? 0 : left);
  long long right = mid - (N - X);
  right = (right < 0 ? 0 : right);
  long long up = mid - (Y - 1);
  up = (up < 0 ? 0 : up);
  long long down = mid - (N - Y);
  down = (down < 0 ? 0 : down);
  long long sub = left * left + right * right + up * up + down * down;
  if (up > (N - X) + 1) sub -= (up - (N - X)) * (up - (N - X) - 1) / 2;
  if (up > (X - 1) + 1) sub -= (up - (X - 1)) * (up - (X - 1) - 1) / 2;
  if (down > (N - X) + 1) sub -= (down - (N - X)) * (down - (N - X) - 1) / 2;
  if (down > (X - 1) + 1) sub -= (down - (X - 1)) * (down - (X - 1) - 1) / 2;
  return e - sub;
}
int main() {
  long long N, X, Y, C;
  cin >> N >> Y >> X >> C;
  long long lo = 0, hi = 3 * N;
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    long long e = funct(mid, X, Y, N);
    if (e < C)
      lo = mid + 1;
    else
      hi = mid;
  }
  cout << lo << endl;
}

### 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;
int arr[5000];
int M[1000010];
int P[1000010];
int ST[1000010];
long long funct(long long mid, long long X, long long Y, long long N) {
  long long e = (mid * mid + (mid + 1) * (mid + 1));
  long long left = mid - (X - 1);
  left = (left < 0 ? 0 : left);
  long long right = mid - (N - X);
  right = (right < 0 ? 0 : right);
  long long up = mid - (Y - 1);
  up = (up < 0 ? 0 : up);
  long long down = mid - (N - Y);
  down = (down < 0 ? 0 : down);
  long long sub = left * left + right * right + up * up + down * down;
  if (up > (N - X) + 1) sub -= (up - (N - X)) * (up - (N - X) - 1) / 2;
  if (up > (X - 1) + 1) sub -= (up - (X - 1)) * (up - (X - 1) - 1) / 2;
  if (down > (N - X) + 1) sub -= (down - (N - X)) * (down - (N - X) - 1) / 2;
  if (down > (X - 1) + 1) sub -= (down - (X - 1)) * (down - (X - 1) - 1) / 2;
  return e - sub;
}
int main() {
  long long N, X, Y, C;
  cin >> N >> Y >> X >> C;
  long long lo = 0, hi = 3 * N;
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    long long e = funct(mid, X, Y, N);
    if (e < C)
      lo = mid + 1;
    else
      hi = mid;
  }
  cout << lo << endl;
}

```