#include <bits/stdc++.h>
using namespace std;
long long n, sx, sy, c;
long long ans;
long long calc(long long s, long long t, long long d) {
  return s <= t ? (s + t) * ((t - s) / d + 1) / 2 : 0;
}
bool legal(long long m) {
  long long res = 0;
  res = calc(1, 2 * m - 1, 2) * 2 + 2 * m + 1;
  long long lef = sy - m, up = sx - m, rig = sy + m, dow = sx + m;
  res -= calc(1, 2 * (1 - lef) - 1, 2) + calc(1, 2 * (1 - up) - 1, 2) +
         calc(1, 2 * (rig - n) - 1, 2) + calc(1, 2 * (dow - n) - 1, 2);
  long long dl = sy, du = sx, dr = n - sy + 1, dd = n - sx + 1;
  res += calc(1, 1 - lef - du, 1) + calc(1, 1 - up - dr, 1) +
         calc(1, rig - n - dd, 1) + calc(1, dow - n - dl, 1);
  return res >= c;
}
signed main() {
  cin >> n >> sx >> sy >> c;
  long long l = 0, r = 2 * n;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (legal(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

### 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, sx, sy, c;
long long ans;
long long calc(long long s, long long t, long long d) {
  return s <= t ? (s + t) * ((t - s) / d + 1) / 2 : 0;
}
bool legal(long long m) {
  long long res = 0;
  res = calc(1, 2 * m - 1, 2) * 2 + 2 * m + 1;
  long long lef = sy - m, up = sx - m, rig = sy + m, dow = sx + m;
  res -= calc(1, 2 * (1 - lef) - 1, 2) + calc(1, 2 * (1 - up) - 1, 2) +
         calc(1, 2 * (rig - n) - 1, 2) + calc(1, 2 * (dow - n) - 1, 2);
  long long dl = sy, du = sx, dr = n - sy + 1, dd = n - sx + 1;
  res += calc(1, 1 - lef - du, 1) + calc(1, 1 - up - dr, 1) +
         calc(1, rig - n - dd, 1) + calc(1, dow - n - dl, 1);
  return res >= c;
}
signed main() {
  cin >> n >> sx >> sy >> c;
  long long l = 0, r = 2 * n;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (legal(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

```