#include <bits/stdc++.h>
using namespace std;
long long S(long long a, long long b) {
  return ((a + b) & 1) ? (b - a + 1) / 2 * (a + b) : (a + b) / 2 * (b - a + 1);
}
long long c(long long a, long long b, long long k) {
  if (!a || !b || !k) return 0;
  if (k > a) {
    if (k - a >= b) return a * b;
    return a * (k - a) + S(a - min(a - 1, (b - k + a - 1)), a);
  } else
    return S(k - min(k - 1, b - 1), k);
}
int main() {
  long long n, x, y, C;
  cin >> n >> x >> y >> C;
  --x;
  --y;
  long long l = 0, r = C - 1;
  while (l < r) {
    long long m = (l + r) / 2;
    long long f = 1;
    long long U = (x >= m) ? m : x;
    long long L = (y >= m) ? m : y;
    long long R = (y + m < n) ? m : (n - y - 1);
    long long D = (x + m < n) ? m : (n - x - 1), ul, ur, dl, dr;
    f += U + L + R + D;
    ul = c(U, L, m - 1);
    ur = c(U, R, m - 1);
    dl = c(D, L, m - 1);
    dr = c(D, R, m - 1);
    f += c(U, L, m - 1) + c(U, R, m - 1) + c(D, L, m - 1) + c(D, R, m - 1);
    if (f >= C)
      r = m;
    else
      l = m + 1;
  }
  cout << l;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long S(long long a, long long b) {
  return ((a + b) & 1) ? (b - a + 1) / 2 * (a + b) : (a + b) / 2 * (b - a + 1);
}
long long c(long long a, long long b, long long k) {
  if (!a || !b || !k) return 0;
  if (k > a) {
    if (k - a >= b) return a * b;
    return a * (k - a) + S(a - min(a - 1, (b - k + a - 1)), a);
  } else
    return S(k - min(k - 1, b - 1), k);
}
int main() {
  long long n, x, y, C;
  cin >> n >> x >> y >> C;
  --x;
  --y;
  long long l = 0, r = C - 1;
  while (l < r) {
    long long m = (l + r) / 2;
    long long f = 1;
    long long U = (x >= m) ? m : x;
    long long L = (y >= m) ? m : y;
    long long R = (y + m < n) ? m : (n - y - 1);
    long long D = (x + m < n) ? m : (n - x - 1), ul, ur, dl, dr;
    f += U + L + R + D;
    ul = c(U, L, m - 1);
    ur = c(U, R, m - 1);
    dl = c(D, L, m - 1);
    dr = c(D, R, m - 1);
    f += c(U, L, m - 1) + c(U, R, m - 1) + c(D, L, m - 1) + c(D, R, m - 1);
    if (f >= C)
      r = m;
    else
      l = m + 1;
  }
  cout << l;
  return 0;
}

```