#include <bits/stdc++.h>
int main() {
  long long n, m;
  int x, y;
  int i, j;
  long long high, low, mid;
  scanf("%I64d%d%d%I64d", &n, &x, &y, &m);
  high = 2 * n;
  low = 0;
  while (low < high) {
    mid = (low + high) / 2;
    long long tot = 0;
    if (x - mid >= 1) {
      if (y - mid >= 1)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - y + 1) * (mid - y + 2) / 2;
    } else {
      if (y - mid >= 1)
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - x + 1) * (mid - x + 2) / 2;
      else {
        if (x - 1 + y - 1 <= mid)
          tot = tot + x * y;
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid - y + 1) * (mid - y + 2) / 2 -
                (mid - x + 1) * (mid - x + 2) / 2;
      }
    }
    if (x - mid >= 1) {
      if (y + mid <= n)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + y - n) * (mid + y - n + 1) / 2;
    } else {
      if (y + mid <= n)
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - x + 1) * (mid - x + 2) / 2;
      else {
        if (x - 1 + n - y <= mid)
          tot = tot + x * (n - y + 1);
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid + y - n) * (mid + y - n + 1) / 2 -
                (mid - x + 1) * (mid - x + 2) / 2;
      }
    }
    if (x + mid <= n) {
      if (y - mid >= 1)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - y + 1) * (mid - y + 2) / 2;
    } else {
      if (y - mid >= 1)
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + x - n) * (mid + x - n + 1) / 2;
      else {
        if (n - x + y - 1 <= mid)
          tot = tot + (n - x + 1) * y;
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid - y + 1) * (mid - y + 2) / 2 -
                (mid + x - n) * (mid + x - n + 1) / 2;
      }
    }
    if (x + mid <= n) {
      if (y + mid <= n)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + y - n) * (mid + y - n + 1) / 2;
    } else {
      if (y + mid <= n)
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + x - n) * (mid + x - n + 1) / 2;
      else {
        if (n - x + n - y <= mid)
          tot = tot + (n - x + 1) * (n - y + 1);
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid + y - n) * (mid + y - n + 1) / 2 -
                (mid + x - n) * (mid + x - n + 1) / 2;
      }
    }
    if (x - mid >= 1)
      tot = tot - (mid + 1);
    else
      tot = tot - x;
    if (x + mid <= n)
      tot = tot - (mid + 1);
    else
      tot = tot - (n - x + 1);
    if (y - mid >= 1)
      tot = tot - (mid + 1);
    else
      tot = tot - y;
    if (y + mid <= n)
      tot = tot - (mid + 1);
    else
      tot = tot - (n - y + 1);
    tot++;
    if (tot < m)
      low = mid + 1;
    else
      high = mid;
  }
  printf("%I64d\n", low);
  return 0;
}

### Prompt

Your challenge is to write 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>
int main() {
  long long n, m;
  int x, y;
  int i, j;
  long long high, low, mid;
  scanf("%I64d%d%d%I64d", &n, &x, &y, &m);
  high = 2 * n;
  low = 0;
  while (low < high) {
    mid = (low + high) / 2;
    long long tot = 0;
    if (x - mid >= 1) {
      if (y - mid >= 1)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - y + 1) * (mid - y + 2) / 2;
    } else {
      if (y - mid >= 1)
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - x + 1) * (mid - x + 2) / 2;
      else {
        if (x - 1 + y - 1 <= mid)
          tot = tot + x * y;
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid - y + 1) * (mid - y + 2) / 2 -
                (mid - x + 1) * (mid - x + 2) / 2;
      }
    }
    if (x - mid >= 1) {
      if (y + mid <= n)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + y - n) * (mid + y - n + 1) / 2;
    } else {
      if (y + mid <= n)
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - x + 1) * (mid - x + 2) / 2;
      else {
        if (x - 1 + n - y <= mid)
          tot = tot + x * (n - y + 1);
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid + y - n) * (mid + y - n + 1) / 2 -
                (mid - x + 1) * (mid - x + 2) / 2;
      }
    }
    if (x + mid <= n) {
      if (y - mid >= 1)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot =
            tot + (mid + 1) * (mid + 2) / 2 - (mid - y + 1) * (mid - y + 2) / 2;
    } else {
      if (y - mid >= 1)
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + x - n) * (mid + x - n + 1) / 2;
      else {
        if (n - x + y - 1 <= mid)
          tot = tot + (n - x + 1) * y;
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid - y + 1) * (mid - y + 2) / 2 -
                (mid + x - n) * (mid + x - n + 1) / 2;
      }
    }
    if (x + mid <= n) {
      if (y + mid <= n)
        tot = tot + (mid + 1) * (mid + 2) / 2;
      else
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + y - n) * (mid + y - n + 1) / 2;
    } else {
      if (y + mid <= n)
        tot = tot + (mid + 1) * (mid + 2) / 2 -
              (mid + x - n) * (mid + x - n + 1) / 2;
      else {
        if (n - x + n - y <= mid)
          tot = tot + (n - x + 1) * (n - y + 1);
        else
          tot = tot + (mid + 1) * (mid + 2) / 2 -
                (mid + y - n) * (mid + y - n + 1) / 2 -
                (mid + x - n) * (mid + x - n + 1) / 2;
      }
    }
    if (x - mid >= 1)
      tot = tot - (mid + 1);
    else
      tot = tot - x;
    if (x + mid <= n)
      tot = tot - (mid + 1);
    else
      tot = tot - (n - x + 1);
    if (y - mid >= 1)
      tot = tot - (mid + 1);
    else
      tot = tot - y;
    if (y + mid <= n)
      tot = tot - (mid + 1);
    else
      tot = tot - (n - y + 1);
    tot++;
    if (tot < m)
      low = mid + 1;
    else
      high = mid;
  }
  printf("%I64d\n", low);
  return 0;
}

```