#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long computeOddRec(long long len) { return len * (len - 1) + len; }
long long computeSeries(long long len) { return len * (len + 1) / 2; }
long long up(long long s) {
  long long len = s - x + 1;
  if (len <= 0) return 0;
  return computeOddRec(len);
}
long long down(long long s) {
  long long len = s - (n - x);
  if (len <= 0) return 0;
  return computeOddRec(len);
}
long long right(long long s) {
  long long len = s - (n - y);
  if (len <= 0) return 0;
  long long sum = computeOddRec(len);
  if (len - x > 0) sum -= computeSeries(len - x);
  if (len - (n - x + 1) > 0) sum -= computeSeries(len - (n - x + 1));
  return sum;
}
long long left(long long s) {
  long long len = s - y + 1;
  if (len <= 0) return 0;
  long long sum = computeOddRec(len);
  if (len - x > 0) sum -= computeSeries(len - x);
  if (len - (n - x + 1) > 0) sum -= computeSeries(len - (n - x + 1));
  return sum;
}
bool ok(long long s) {
  if (s == 4) s = s;
  long long sum = 4 * (s * (s + 1) / 2) + 1;
  sum -= up(s) + down(s) + right(s) + left(s);
  assert(sum > 0);
  if (sum >= c) return true;
  return false;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &x, &y, &c);
  long long l = 0, r = 1e9 + 1, mid, minn = n * n + 1;
  while (l < r) {
    mid = (l + r) / 2;
    if (ok(mid)) {
      minn = min(minn, mid);
      r = mid;
    } else
      l = mid + 1;
  }
  assert(minn != n * n + 1);
  printf("%lld", minn);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, x, y, c;
long long computeOddRec(long long len) { return len * (len - 1) + len; }
long long computeSeries(long long len) { return len * (len + 1) / 2; }
long long up(long long s) {
  long long len = s - x + 1;
  if (len <= 0) return 0;
  return computeOddRec(len);
}
long long down(long long s) {
  long long len = s - (n - x);
  if (len <= 0) return 0;
  return computeOddRec(len);
}
long long right(long long s) {
  long long len = s - (n - y);
  if (len <= 0) return 0;
  long long sum = computeOddRec(len);
  if (len - x > 0) sum -= computeSeries(len - x);
  if (len - (n - x + 1) > 0) sum -= computeSeries(len - (n - x + 1));
  return sum;
}
long long left(long long s) {
  long long len = s - y + 1;
  if (len <= 0) return 0;
  long long sum = computeOddRec(len);
  if (len - x > 0) sum -= computeSeries(len - x);
  if (len - (n - x + 1) > 0) sum -= computeSeries(len - (n - x + 1));
  return sum;
}
bool ok(long long s) {
  if (s == 4) s = s;
  long long sum = 4 * (s * (s + 1) / 2) + 1;
  sum -= up(s) + down(s) + right(s) + left(s);
  assert(sum > 0);
  if (sum >= c) return true;
  return false;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &x, &y, &c);
  long long l = 0, r = 1e9 + 1, mid, minn = n * n + 1;
  while (l < r) {
    mid = (l + r) / 2;
    if (ok(mid)) {
      minn = min(minn, mid);
      r = mid;
    } else
      l = mid + 1;
  }
  assert(minn != n * n + 1);
  printf("%lld", minn);
  return 0;
}

```