#include <bits/stdc++.h>
using namespace std;
int n, x, y, c, l, r, mid;
long long P(long long x) {
  if (x <= 0) return 0;
  return x * x;
}
long long Q(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
bool check(long long p) {
  long long res = 0;
  res = (p + 1) * (p + 1) + p * p;
  res = res - P(1 - (x - p)) - P(x + p - n) - P(y + p - n) - P(1 - (y - p));
  res = res + Q(p - (x - 1) - (y - 1) - 1) + Q(p - (x - 1) - (n - y) - 1) +
        Q(p - (n - x) - (y - 1) - 1) + Q(p - (n - x) - (n - y) - 1);
  return res >= c;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  l = 0;
  r = 100000;
  while (l <= r) {
    mid = l + r >> 1;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", r + 1);
}

### 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 n, x, y, c, l, r, mid;
long long P(long long x) {
  if (x <= 0) return 0;
  return x * x;
}
long long Q(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
bool check(long long p) {
  long long res = 0;
  res = (p + 1) * (p + 1) + p * p;
  res = res - P(1 - (x - p)) - P(x + p - n) - P(y + p - n) - P(1 - (y - p));
  res = res + Q(p - (x - 1) - (y - 1) - 1) + Q(p - (x - 1) - (n - y) - 1) +
        Q(p - (n - x) - (y - 1) - 1) + Q(p - (n - x) - (n - y) - 1);
  return res >= c;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  l = 0;
  r = 100000;
  while (l <= r) {
    mid = l + r >> 1;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%d\n", r + 1);
}

```