#include <bits/stdc++.h>
using namespace std;
int stmp;
const int MAX = 100000;
const int INF = 1000000001;
int n, x, y, c;
int main(int argc, char *argv[]) {
  stmp = scanf("%d %d %d %d", &n, &x, &y, &c);
  --c;
  int k = 0;
  while (c > 0) {
    ++k;
    int cnt = 4 * k;
    cnt -= max(0, 2 * (y + k - n) - 1);
    cnt -= max(0, 2 * (x + k - n) - 1);
    cnt -= max(0, 2 * (1 - (y - k)) - 1);
    cnt -= max(0, 2 * (1 - (x - k)) - 1);
    int l;
    l = n - x + n - y + 1;
    cnt += max(0, k - l);
    l = n - x + y;
    cnt += max(0, k - l);
    l = x + n - y;
    cnt += max(0, k - l);
    l = x + y - 1;
    cnt += max(0, k - l);
    c -= cnt;
  }
  printf("%d\n", k);
  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 stmp;
const int MAX = 100000;
const int INF = 1000000001;
int n, x, y, c;
int main(int argc, char *argv[]) {
  stmp = scanf("%d %d %d %d", &n, &x, &y, &c);
  --c;
  int k = 0;
  while (c > 0) {
    ++k;
    int cnt = 4 * k;
    cnt -= max(0, 2 * (y + k - n) - 1);
    cnt -= max(0, 2 * (x + k - n) - 1);
    cnt -= max(0, 2 * (1 - (y - k)) - 1);
    cnt -= max(0, 2 * (1 - (x - k)) - 1);
    int l;
    l = n - x + n - y + 1;
    cnt += max(0, k - l);
    l = n - x + y;
    cnt += max(0, k - l);
    l = x + n - y;
    cnt += max(0, k - l);
    l = x + y - 1;
    cnt += max(0, k - l);
    c -= cnt;
  }
  printf("%d\n", k);
  return 0;
}

```