#include <bits/stdc++.h>
using namespace std;
long long cal(long long s) {
  long long ret = (1LL + s) * s / 2;
  return 1LL + ret * 4LL;
}
long long cal2(long long s) {
  long long tm = (s * 2LL - 1LL);
  long long ret = (tm + 1) * s / 2LL;
  return ret;
}
long long cal3(long long s) {
  long long ret = (s + 1) * s / 2;
  return ret;
}
int main() {
  int n, x, y;
  long long c;
  scanf("%d%d%d%I64d", &n, &x, &y, &c);
  long long tmp = 1;
  int ans = 0;
  while (tmp < c) {
    ans++;
    int mid = ans;
    tmp = cal(mid);
    if (x + mid > n) tmp -= cal2(x + mid - n);
    if (x - mid < 1) tmp -= cal2(mid - x + 1);
    if (y + mid > n) tmp -= cal2(y + mid - n);
    if (y - mid < 1) tmp -= cal2(mid - y + 1);
    if (x + y <= mid) tmp += cal3(mid - (x + y) + 1);
    if (x + n + 1 - y <= mid) tmp += cal3(mid - (x + n + 1 - y) + 1);
    if (n + 1 - x + y <= mid) tmp += cal3(mid - (n + 1 - x + y) + 1);
    if (n + 1 - x + n + 1 - y <= mid)
      tmp += cal3(mid - (n + 1 - x + n + 1 - y) + 1);
  }
  printf("%d\n", ans);
}

### Prompt

Your task is to create 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;
long long cal(long long s) {
  long long ret = (1LL + s) * s / 2;
  return 1LL + ret * 4LL;
}
long long cal2(long long s) {
  long long tm = (s * 2LL - 1LL);
  long long ret = (tm + 1) * s / 2LL;
  return ret;
}
long long cal3(long long s) {
  long long ret = (s + 1) * s / 2;
  return ret;
}
int main() {
  int n, x, y;
  long long c;
  scanf("%d%d%d%I64d", &n, &x, &y, &c);
  long long tmp = 1;
  int ans = 0;
  while (tmp < c) {
    ans++;
    int mid = ans;
    tmp = cal(mid);
    if (x + mid > n) tmp -= cal2(x + mid - n);
    if (x - mid < 1) tmp -= cal2(mid - x + 1);
    if (y + mid > n) tmp -= cal2(y + mid - n);
    if (y - mid < 1) tmp -= cal2(mid - y + 1);
    if (x + y <= mid) tmp += cal3(mid - (x + y) + 1);
    if (x + n + 1 - y <= mid) tmp += cal3(mid - (x + n + 1 - y) + 1);
    if (n + 1 - x + y <= mid) tmp += cal3(mid - (n + 1 - x + y) + 1);
    if (n + 1 - x + n + 1 - y <= mid)
      tmp += cal3(mid - (n + 1 - x + n + 1 - y) + 1);
  }
  printf("%d\n", ans);
}

```