#include <bits/stdc++.h>
const int inf = (1ll << 31) - 1;
const int maxn = (int)5e5 + 10;
const int mod = (int)1e9 + 7;
using namespace std;
int n, m, x, y, c;
bool ok(int cur) {
  int x1 = max(x - cur, 1);
  int x2 = min(x + cur, n);
  long long ans = 0;
  for (int i = x1; i <= x2; i++) {
    int z1 = cur - abs(i - x);
    int k1 = min(y + z1, n);
    int k2 = max(y - z1, 1);
    ans += k1 - k2 + 1;
  }
  return ans >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = maxn;
  int ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (ok(mid)) {
      r = mid - 1;
      ans = mid;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

### 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>
const int inf = (1ll << 31) - 1;
const int maxn = (int)5e5 + 10;
const int mod = (int)1e9 + 7;
using namespace std;
int n, m, x, y, c;
bool ok(int cur) {
  int x1 = max(x - cur, 1);
  int x2 = min(x + cur, n);
  long long ans = 0;
  for (int i = x1; i <= x2; i++) {
    int z1 = cur - abs(i - x);
    int k1 = min(y + z1, n);
    int k2 = max(y - z1, 1);
    ans += k1 - k2 + 1;
  }
  return ans >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = maxn;
  int ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (ok(mid)) {
      r = mid - 1;
      ans = mid;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

```