#include <bits/stdc++.h>
long long n, x, y, c, l, r, mid;
long long s1, s2, s3, s4, s5, s6, s7, s8, s9;
void init() { scanf("%lld%lld%lld%lld", &n, &x, &y, &c); }
long long Get(long long x) { return x * x; }
long long G(long long x) { return x * (x + 1) / 2; }
bool check(long long p) {
  long long s1 = 2 * p * (p + 1) + 1;
  s2 = s3 = s4 = s5 = s6 = s7 = s8 = s9 = 0;
  if (x + p > n) s2 = Get(x + p - n);
  if (y + p > n) s3 = Get(y + p - n);
  if (x - p < 1) s4 = Get(p - x + 1);
  if (y - p < 1) s5 = Get(p - y + 1);
  if (x + y + p > n * 2 + 1) s6 = G(x + y + p - n * 2 - 1);
  if (x + p - n - y + 1 > 1) s7 = G(x + p - n - y + 1 - 1);
  if (y + p - n - x + 1 > 1) s8 = G(y + p - n - x + 1 - 1);
  if (p - x + 1 - y + 1 > 1) s9 = G(p - x + 1 - y + 1 - 1);
  return s1 - s2 - s3 - s4 - s5 + s6 + s7 + s8 + s9 >= c;
}
void work() {
  l = 0;
  r = n * 2;
  while (l < r) {
    mid = l + r >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  printf("%lld", r);
}
int main() {
  init();
  work();
  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>
long long n, x, y, c, l, r, mid;
long long s1, s2, s3, s4, s5, s6, s7, s8, s9;
void init() { scanf("%lld%lld%lld%lld", &n, &x, &y, &c); }
long long Get(long long x) { return x * x; }
long long G(long long x) { return x * (x + 1) / 2; }
bool check(long long p) {
  long long s1 = 2 * p * (p + 1) + 1;
  s2 = s3 = s4 = s5 = s6 = s7 = s8 = s9 = 0;
  if (x + p > n) s2 = Get(x + p - n);
  if (y + p > n) s3 = Get(y + p - n);
  if (x - p < 1) s4 = Get(p - x + 1);
  if (y - p < 1) s5 = Get(p - y + 1);
  if (x + y + p > n * 2 + 1) s6 = G(x + y + p - n * 2 - 1);
  if (x + p - n - y + 1 > 1) s7 = G(x + p - n - y + 1 - 1);
  if (y + p - n - x + 1 > 1) s8 = G(y + p - n - x + 1 - 1);
  if (p - x + 1 - y + 1 > 1) s9 = G(p - x + 1 - y + 1 - 1);
  return s1 - s2 - s3 - s4 - s5 + s6 + s7 + s8 + s9 >= c;
}
void work() {
  l = 0;
  r = n * 2;
  while (l < r) {
    mid = l + r >> 1;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  printf("%lld", r);
}
int main() {
  init();
  work();
  return 0;
}

```