#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
long long get_center(long long n) { return 2 * n * (n + 1) + 1; }
long long get_side(long long n) {
  if (n < 0) return 0;
  return (n + 1) * (n + 1);
}
long long get_corner(long long n) {
  if (n < 0) return 0;
  return n * (n + 3) / 2 + 1;
}
int n, x, y, c;
long long f(int k) {
  long long ret = 0;
  ret += get_center(k);
  ret -= get_side(0 - (x - k));
  ret -= get_side((x + k) - (n + 1));
  ret -= get_side(0 - (y - k));
  ret -= get_side((y + k) - (n + 1));
  ret += get_corner(k - (x - 0 + y - 0));
  ret += get_corner(k - (x - 0 + n + 1 - y));
  ret += get_corner(k - (n + 1 - x + y - 0));
  ret += get_corner(k - (n + 1 - x + n + 1 - y));
  return ret;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int l = 0, r = n + n;
  while (l < r) {
    int m = l + r >> 1;
    if (f(m) < c)
      l = m + 1;
    else
      r = m;
  }
  printf("%d\n", r);
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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>
using namespace std;
const int N = 1e5 + 5;
long long get_center(long long n) { return 2 * n * (n + 1) + 1; }
long long get_side(long long n) {
  if (n < 0) return 0;
  return (n + 1) * (n + 1);
}
long long get_corner(long long n) {
  if (n < 0) return 0;
  return n * (n + 3) / 2 + 1;
}
int n, x, y, c;
long long f(int k) {
  long long ret = 0;
  ret += get_center(k);
  ret -= get_side(0 - (x - k));
  ret -= get_side((x + k) - (n + 1));
  ret -= get_side(0 - (y - k));
  ret -= get_side((y + k) - (n + 1));
  ret += get_corner(k - (x - 0 + y - 0));
  ret += get_corner(k - (x - 0 + n + 1 - y));
  ret += get_corner(k - (n + 1 - x + y - 0));
  ret += get_corner(k - (n + 1 - x + n + 1 - y));
  return ret;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int l = 0, r = n + n;
  while (l < r) {
    int m = l + r >> 1;
    if (f(m) < c)
      l = m + 1;
    else
      r = m;
  }
  printf("%d\n", r);
  return 0;
}

```