#include <bits/stdc++.h>
using namespace std;
const double PI = 2.0 * acos(0.0);
const double EPS = 1e-7;
map<int, vector<int> > M;
int main() {
  int n, x, y, c;
  scanf("%d %d %d %d", &n, &x, &y, &c);
  int arrive_time[4] = {x + y, x + (n + 1 - y), y + (n + 1 - x),
                        (n + 1 - x) + (n + 1 - y)};
  for (int t = 0; true; t++) {
    int cell = 2 * t * t + 2 * t + 1;
    for (int dir = 0; dir < 4; dir++)
      if (t >= arrive_time[dir]) {
        int newt = t - arrive_time[dir] + 1;
        cell += newt * (newt + 1) / 2;
      }
    if (x <= t) {
      int newt = t - x + 1;
      cell -= newt * newt;
    }
    if (y <= t) {
      int newt = t - y + 1;
      cell -= newt * newt;
    }
    if (x >= n + 1 - t) {
      int newt = x - (n + 1 - t) + 1;
      cell -= newt * newt;
    }
    if (y >= n + 1 - t) {
      int newt = y - (n + 1 - t) + 1;
      cell -= newt * newt;
    }
    if (cell >= c) {
      printf("%d\n", t);
      break;
    }
  }
  return 0;
}

### Prompt

Generate 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;
const double PI = 2.0 * acos(0.0);
const double EPS = 1e-7;
map<int, vector<int> > M;
int main() {
  int n, x, y, c;
  scanf("%d %d %d %d", &n, &x, &y, &c);
  int arrive_time[4] = {x + y, x + (n + 1 - y), y + (n + 1 - x),
                        (n + 1 - x) + (n + 1 - y)};
  for (int t = 0; true; t++) {
    int cell = 2 * t * t + 2 * t + 1;
    for (int dir = 0; dir < 4; dir++)
      if (t >= arrive_time[dir]) {
        int newt = t - arrive_time[dir] + 1;
        cell += newt * (newt + 1) / 2;
      }
    if (x <= t) {
      int newt = t - x + 1;
      cell -= newt * newt;
    }
    if (y <= t) {
      int newt = t - y + 1;
      cell -= newt * newt;
    }
    if (x >= n + 1 - t) {
      int newt = x - (n + 1 - t) + 1;
      cell -= newt * newt;
    }
    if (y >= n + 1 - t) {
      int newt = y - (n + 1 - t) + 1;
      cell -= newt * newt;
    }
    if (cell >= c) {
      printf("%d\n", t);
      break;
    }
  }
  return 0;
}

```