#include <bits/stdc++.h>
using namespace std;
const int INF = 1500000002;
const long long INFLL = 100000000000000000ll;
const long long MOD = 1000000007;
long long N, X, Y, C;
long long sum(long long a, long long b) {
  if (a > b) return 0;
  long long q = b - a + 1;
  long long ret = q * (q + 1) / 2;
  ret += (a - 1) * q;
  return ret;
}
long long simul(long long steps) {
  long long ret = steps * (steps + 1) * 2 + 1;
  long long l = min(X - 1, steps);
  long long r = min(N - X, steps);
  long long up = max(1 - (Y - steps), 0ll);
  long long down = max(Y + steps - N, 0ll);
  long long a = max(up - l, 0ll);
  long long b = up - 1;
  ret -= sum(a, b);
  a = max(up - r, 0ll);
  b = up - 1;
  ret -= sum(a, b);
  a = max(down - l, 0ll);
  b = down - 1;
  ret -= sum(a, b);
  a = max(down - r, 0ll);
  b = down - 1;
  ret -= sum(a, b);
  ret -= down;
  ret -= up;
  long long left = max(1 - (X - steps), 0ll);
  ret -= left * left;
  long long right = max(X + steps - N, 0ll);
  ret -= right * right;
  return ret;
}
int main() {
  cin >> N >> X >> Y >> C;
  long long a = 0, b = 500000000;
  while (b - a > 1) {
    long long p = (a + b) / 2;
    long long have = simul(p);
    if (have >= C)
      b = p;
    else
      a = p;
  }
  if (simul(a) >= C)
    cout << a;
  else
    cout << b;
}

### 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;
const int INF = 1500000002;
const long long INFLL = 100000000000000000ll;
const long long MOD = 1000000007;
long long N, X, Y, C;
long long sum(long long a, long long b) {
  if (a > b) return 0;
  long long q = b - a + 1;
  long long ret = q * (q + 1) / 2;
  ret += (a - 1) * q;
  return ret;
}
long long simul(long long steps) {
  long long ret = steps * (steps + 1) * 2 + 1;
  long long l = min(X - 1, steps);
  long long r = min(N - X, steps);
  long long up = max(1 - (Y - steps), 0ll);
  long long down = max(Y + steps - N, 0ll);
  long long a = max(up - l, 0ll);
  long long b = up - 1;
  ret -= sum(a, b);
  a = max(up - r, 0ll);
  b = up - 1;
  ret -= sum(a, b);
  a = max(down - l, 0ll);
  b = down - 1;
  ret -= sum(a, b);
  a = max(down - r, 0ll);
  b = down - 1;
  ret -= sum(a, b);
  ret -= down;
  ret -= up;
  long long left = max(1 - (X - steps), 0ll);
  ret -= left * left;
  long long right = max(X + steps - N, 0ll);
  ret -= right * right;
  return ret;
}
int main() {
  cin >> N >> X >> Y >> C;
  long long a = 0, b = 500000000;
  while (b - a > 1) {
    long long p = (a + b) / 2;
    long long have = simul(p);
    if (have >= C)
      b = p;
    else
      a = p;
  }
  if (simul(a) >= C)
    cout << a;
  else
    cout << b;
}

```