#include <bits/stdc++.h>
using namespace std;
int n;
int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
bool out(pair<int, int> p) {
  return p.first > n || p.first < 1 || p.second > n || p.second < 1;
}
void apply(int i, pair<int, int> &p) {
  p.first += dir[i][0];
  p.second += dir[i][1];
}
int main() {
  int X, Y, c, t = 0;
  cin >> n >> X >> Y >> c;
  long long sum = 1, res[4] = {1, 1, 1, 1};
  pair<pair<int, int>, pair<int, int> > dia[4];
  for (int i = 0; i < 4; ++i) {
    pair<int, int> tmp(X + dir[i][0], Y + dir[i][1]);
    dia[i] = pair<pair<int, int>, pair<int, int> >(tmp, tmp);
    if (out(tmp)) res[i] = 0;
  }
  while (sum < c) {
    for (int i = 0; i < 4; ++i) {
      if (res[i] <= 0) continue;
      sum += res[i];
      res[i]++;
      apply(i, dia[i].first);
      apply((i + 1) % 4, dia[i].second);
      if (out(dia[i].first)) {
        apply((i + 1) % 4, dia[i].first);
        apply((i + 2) % 4, dia[i].first);
        res[i]--;
      }
      if (out(dia[i].second)) {
        apply((i + 1) % 4, dia[i].second);
        apply((i + 2) % 4, dia[i].second);
        res[i]--;
      }
    }
    ++t;
  }
  cout << t << endl;
  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;
int n;
int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
bool out(pair<int, int> p) {
  return p.first > n || p.first < 1 || p.second > n || p.second < 1;
}
void apply(int i, pair<int, int> &p) {
  p.first += dir[i][0];
  p.second += dir[i][1];
}
int main() {
  int X, Y, c, t = 0;
  cin >> n >> X >> Y >> c;
  long long sum = 1, res[4] = {1, 1, 1, 1};
  pair<pair<int, int>, pair<int, int> > dia[4];
  for (int i = 0; i < 4; ++i) {
    pair<int, int> tmp(X + dir[i][0], Y + dir[i][1]);
    dia[i] = pair<pair<int, int>, pair<int, int> >(tmp, tmp);
    if (out(tmp)) res[i] = 0;
  }
  while (sum < c) {
    for (int i = 0; i < 4; ++i) {
      if (res[i] <= 0) continue;
      sum += res[i];
      res[i]++;
      apply(i, dia[i].first);
      apply((i + 1) % 4, dia[i].second);
      if (out(dia[i].first)) {
        apply((i + 1) % 4, dia[i].first);
        apply((i + 2) % 4, dia[i].first);
        res[i]--;
      }
      if (out(dia[i].second)) {
        apply((i + 1) % 4, dia[i].second);
        apply((i + 2) % 4, dia[i].second);
        res[i]--;
      }
    }
    ++t;
  }
  cout << t << endl;
  return 0;
}

```