#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
long long sum(int n) { return 1LL * (n) * (n + 1) / 2; }
long long get(int x, int y, int m) {
  long long ret = 0;
  if (x < 1 || y < 1 || m <= 0) return 0;
  int addx = max(m - x, 0), addy = max(m - y, 0);
  ret += 1LL * (addx) * (addx + 1) / 2 + 1LL * (addy) * (addy + 1) / 2;
  ret -= sum(max(m - x - y, 0));
  return sum(m) - ret;
}
int main() {
  int n, x, y, c;
  cin >> n >> x >> y >> c;
  int l = 0, r = 2 * n + 1;
  while (r - l > 1) {
    int m = (l + r) >> 1;
    long long sum = get(x, y, m);
    sum += get(x, n - y, m - 1);
    sum += get(n - x, y, m - 1);
    sum += get(n - x, n - y, m - 2);
    if (sum >= c)
      r = m;
    else
      l = m;
  }
  cout << r - 1 << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
long long sum(int n) { return 1LL * (n) * (n + 1) / 2; }
long long get(int x, int y, int m) {
  long long ret = 0;
  if (x < 1 || y < 1 || m <= 0) return 0;
  int addx = max(m - x, 0), addy = max(m - y, 0);
  ret += 1LL * (addx) * (addx + 1) / 2 + 1LL * (addy) * (addy + 1) / 2;
  ret -= sum(max(m - x - y, 0));
  return sum(m) - ret;
}
int main() {
  int n, x, y, c;
  cin >> n >> x >> y >> c;
  int l = 0, r = 2 * n + 1;
  while (r - l > 1) {
    int m = (l + r) >> 1;
    long long sum = get(x, y, m);
    sum += get(x, n - y, m - 1);
    sum += get(n - x, y, m - 1);
    sum += get(n - x, n - y, m - 2);
    if (sum >= c)
      r = m;
    else
      l = m;
  }
  cout << r - 1 << endl;
  return 0;
}

```