#include <bits/stdc++.h>
using namespace std;
int n, m, x, y, z, w;
int high, low, mid;
long long k, tmp;
long long solve(int x, int y, int t) {
  int incto = min(x, y);
  int startdec = max(x, y);
  int end = startdec + incto;
  if (t <= incto) {
    long long a = t;
    return (a + 1ll) * (a + 2ll) / 2ll;
  } else if (t > incto && t <= startdec) {
    long long a = incto;
    a = (a + 1ll) * (a + 2ll) / 2ll;
    a += (long long)(t - incto) * (long long)(incto + 1);
    return a;
  } else if (t > startdec && t <= end) {
    long long a = incto;
    long long b = incto;
    a = (a + 1ll) * (a + 2ll) / 2ll;
    a += (long long)(startdec - incto) * (long long)(incto + 1);
    a += b * (b + 1ll) / 2ll;
    long long c = end - t;
    a -= c * (c + 1ll) / 2ll;
    return a;
  } else if (t > end) {
    long long a = incto;
    long long b = incto;
    a = (a + 1ll) * (a + 2ll) / 2ll;
    a += (long long)(startdec - incto) * (long long)(incto + 1);
    a += b * (b + 1ll) / 2ll;
    return a;
  }
}
int main() {
  scanf("%d %d %d %I64d", &n, &x, &y, &k);
  if (k <= 1ll) {
    printf("0\n");
    return 0;
  }
  x--;
  y--;
  high = n * 2;
  low = 0;
  while (high - low > 1) {
    mid = (high + low) / 2;
    tmp = solve(x, y, mid);
    tmp += solve(x, n - y - 1, mid);
    tmp += solve(n - x - 1, y, mid);
    tmp += solve(n - x - 1, n - y - 1, mid);
    tmp -= (long long)min(x, mid);
    tmp -= (long long)min(y, mid);
    tmp -= (long long)min(n - x - 1, mid);
    tmp -= (long long)min(n - y - 1, mid);
    tmp -= 3ll;
    if (tmp >= k)
      high = mid;
    else
      low = mid;
  }
  printf("%d\n", high);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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, m, x, y, z, w;
int high, low, mid;
long long k, tmp;
long long solve(int x, int y, int t) {
  int incto = min(x, y);
  int startdec = max(x, y);
  int end = startdec + incto;
  if (t <= incto) {
    long long a = t;
    return (a + 1ll) * (a + 2ll) / 2ll;
  } else if (t > incto && t <= startdec) {
    long long a = incto;
    a = (a + 1ll) * (a + 2ll) / 2ll;
    a += (long long)(t - incto) * (long long)(incto + 1);
    return a;
  } else if (t > startdec && t <= end) {
    long long a = incto;
    long long b = incto;
    a = (a + 1ll) * (a + 2ll) / 2ll;
    a += (long long)(startdec - incto) * (long long)(incto + 1);
    a += b * (b + 1ll) / 2ll;
    long long c = end - t;
    a -= c * (c + 1ll) / 2ll;
    return a;
  } else if (t > end) {
    long long a = incto;
    long long b = incto;
    a = (a + 1ll) * (a + 2ll) / 2ll;
    a += (long long)(startdec - incto) * (long long)(incto + 1);
    a += b * (b + 1ll) / 2ll;
    return a;
  }
}
int main() {
  scanf("%d %d %d %I64d", &n, &x, &y, &k);
  if (k <= 1ll) {
    printf("0\n");
    return 0;
  }
  x--;
  y--;
  high = n * 2;
  low = 0;
  while (high - low > 1) {
    mid = (high + low) / 2;
    tmp = solve(x, y, mid);
    tmp += solve(x, n - y - 1, mid);
    tmp += solve(n - x - 1, y, mid);
    tmp += solve(n - x - 1, n - y - 1, mid);
    tmp -= (long long)min(x, mid);
    tmp -= (long long)min(y, mid);
    tmp -= (long long)min(n - x - 1, mid);
    tmp -= (long long)min(n - y - 1, mid);
    tmp -= 3ll;
    if (tmp >= k)
      high = mid;
    else
      low = mid;
  }
  printf("%d\n", high);
  return 0;
}

```