#include <bits/stdc++.h>
using namespace std;
int n, c, x, y;
unsigned long long sqr(int num) { return (unsigned long long)num * num; }
unsigned long long getsum(int len) {
  unsigned long long ans = sqr(len + 1) + sqr(len);
  int lx = 1 - (x - len), ly = 1 - (y - len), rx = x + len - n,
      ry = y + len - n, t;
  t = -x + ly;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  t = -x + ry;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  t = x + ly - 1 - n;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  t = x + ry - 1 - n;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  if (lx > 0) ans -= sqr(lx);
  if (ly > 0) ans -= sqr(ly);
  if (rx > 0) ans -= sqr(rx);
  if (ry > 0) ans -= sqr(ry);
  return ans;
}
int findans() {
  int l = 0, r = n * 2 - 1;
  while (l < r) {
    int mid = (l + r) / 2;
    if (getsum(mid) >= (unsigned long long)(c))
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  printf("%d", findans());
  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>
using namespace std;
int n, c, x, y;
unsigned long long sqr(int num) { return (unsigned long long)num * num; }
unsigned long long getsum(int len) {
  unsigned long long ans = sqr(len + 1) + sqr(len);
  int lx = 1 - (x - len), ly = 1 - (y - len), rx = x + len - n,
      ry = y + len - n, t;
  t = -x + ly;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  t = -x + ry;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  t = x + ly - 1 - n;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  t = x + ry - 1 - n;
  if (t > 0) ans += (unsigned long long)t * (t + 1) / 2;
  if (lx > 0) ans -= sqr(lx);
  if (ly > 0) ans -= sqr(ly);
  if (rx > 0) ans -= sqr(rx);
  if (ry > 0) ans -= sqr(ry);
  return ans;
}
int findans() {
  int l = 0, r = n * 2 - 1;
  while (l < r) {
    int mid = (l + r) / 2;
    if (getsum(mid) >= (unsigned long long)(c))
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  printf("%d", findans());
  return 0;
}

```