#include <bits/stdc++.h>
using namespace std;
long long n, c;
long long a[10];
bool check(long long k) {
  if (k <= a[1]) {
    long long tp = 2 * k * (k + 1) + 1;
    return tp >= c;
  }
  if (k <= a[2]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    return tp >= c;
  }
  if (k <= a[3]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    tp -= (k - a[2]) * (k - a[2]);
    if (a[1] + a[2] + 2 <= k)
      tp += (k - a[1] - a[2]) * (k - a[1] - a[2] - 1) / 2;
    return tp >= c;
  }
  if (k <= a[4]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    tp -= (k - a[2]) * (k - a[2]);
    tp -= (k - a[3]) * (k - a[3]);
    if (a[1] + a[2] + 2 <= k)
      tp += (k - a[1] - a[2]) * (k - a[1] - a[2] - 1) / 2;
    if (a[3] + a[1] + 2 <= k)
      tp += (k - a[3] - a[1]) * (k - a[3] - a[1] - 1) / 2;
    return tp >= c;
  }
  if (k > a[4]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    tp -= (k - a[2]) * (k - a[2]);
    tp -= (k - a[3]) * (k - a[3]);
    tp -= (k - a[4]) * (k - a[4]);
    if (a[1] + a[2] + 2 <= k)
      tp += (k - a[1] - a[2]) * (k - a[1] - a[2] - 1) / 2;
    if (a[4] + a[2] + 2 <= k)
      tp += (k - a[4] - a[2]) * (k - a[4] - a[2] - 1) / 2;
    if (a[1] + a[3] + 2 <= k)
      tp += (k - a[1] - a[3]) * (k - a[1] - a[3] - 1) / 2;
    if (a[3] + a[4] + 2 <= k)
      tp += (k - a[3] - a[4]) * (k - a[3] - a[4] - 1) / 2;
    return tp >= c;
  }
}
void solve() {
  long long x, y;
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  a[1] = x - 1;
  a[2] = y - 1;
  a[3] = n - x;
  a[4] = n - y;
  sort(a + 1, a + 5);
  long long l = 0, r = 2 * n;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (!check(mid))
      l = mid + 1;
    else
      r = mid;
  }
  printf("%d\n", l);
}
int main() {
  solve();
  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;
long long n, c;
long long a[10];
bool check(long long k) {
  if (k <= a[1]) {
    long long tp = 2 * k * (k + 1) + 1;
    return tp >= c;
  }
  if (k <= a[2]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    return tp >= c;
  }
  if (k <= a[3]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    tp -= (k - a[2]) * (k - a[2]);
    if (a[1] + a[2] + 2 <= k)
      tp += (k - a[1] - a[2]) * (k - a[1] - a[2] - 1) / 2;
    return tp >= c;
  }
  if (k <= a[4]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    tp -= (k - a[2]) * (k - a[2]);
    tp -= (k - a[3]) * (k - a[3]);
    if (a[1] + a[2] + 2 <= k)
      tp += (k - a[1] - a[2]) * (k - a[1] - a[2] - 1) / 2;
    if (a[3] + a[1] + 2 <= k)
      tp += (k - a[3] - a[1]) * (k - a[3] - a[1] - 1) / 2;
    return tp >= c;
  }
  if (k > a[4]) {
    long long tp = 2 * k * (k + 1) + 1;
    tp -= (k - a[1]) * (k - a[1]);
    tp -= (k - a[2]) * (k - a[2]);
    tp -= (k - a[3]) * (k - a[3]);
    tp -= (k - a[4]) * (k - a[4]);
    if (a[1] + a[2] + 2 <= k)
      tp += (k - a[1] - a[2]) * (k - a[1] - a[2] - 1) / 2;
    if (a[4] + a[2] + 2 <= k)
      tp += (k - a[4] - a[2]) * (k - a[4] - a[2] - 1) / 2;
    if (a[1] + a[3] + 2 <= k)
      tp += (k - a[1] - a[3]) * (k - a[1] - a[3] - 1) / 2;
    if (a[3] + a[4] + 2 <= k)
      tp += (k - a[3] - a[4]) * (k - a[3] - a[4] - 1) / 2;
    return tp >= c;
  }
}
void solve() {
  long long x, y;
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  a[1] = x - 1;
  a[2] = y - 1;
  a[3] = n - x;
  a[4] = n - y;
  sort(a + 1, a + 5);
  long long l = 0, r = 2 * n;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (!check(mid))
      l = mid + 1;
    else
      r = mid;
  }
  printf("%d\n", l);
}
int main() {
  solve();
  return 0;
}

```