#include <bits/stdc++.h>
using namespace std;
inline long long tr(long long d) { return d * (d + 1) / 2; }
inline long long tr2(long long d) { return 2 * tr(d) - d; }
int n;
long long solve(int y, int x, long long c) {
  long long res = 4 * (tr(c + 1) - c - 1) + 1;
  if (c - (y + 1) >= 0) res -= tr2(c - (y + 1) + 1);
  if (c - (n - y) >= 0) res -= tr2(c - (n - y) + 1);
  if (c - (x + 1) >= 0) res -= tr2(c - (x + 1) + 1);
  if (c - (n - x) >= 0) res -= tr2(c - (n - x) + 1);
  if (c - (y + x + 2) >= 0) res += tr(c - (y + x + 2) + 1);
  if (c - (y + 1 + n - x) >= 0) res += tr(c - (y + 1 + n - x) + 1);
  if (c - (x + 1 + n - y) >= 0) res += tr(c - (x + 1 + n - y) + 1);
  if (c - (n - x + n - y) >= 0) res += tr(c - (n - x + n - y) + 1);
  return res;
}
int main() {
  int y, x, c;
  while (scanf("%d%d%d%d", &n, &y, &x, &c) >= 1) {
    --y, --x;
    int l = -1, r = (int)2e9 + 1;
    while (l + 1 < r) {
      int m = l + (r - l) / 2;
      if (solve(y, x, m) >= (long long)c)
        r = m;
      else
        l = m;
    }
    printf("%d\n", r);
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
inline long long tr(long long d) { return d * (d + 1) / 2; }
inline long long tr2(long long d) { return 2 * tr(d) - d; }
int n;
long long solve(int y, int x, long long c) {
  long long res = 4 * (tr(c + 1) - c - 1) + 1;
  if (c - (y + 1) >= 0) res -= tr2(c - (y + 1) + 1);
  if (c - (n - y) >= 0) res -= tr2(c - (n - y) + 1);
  if (c - (x + 1) >= 0) res -= tr2(c - (x + 1) + 1);
  if (c - (n - x) >= 0) res -= tr2(c - (n - x) + 1);
  if (c - (y + x + 2) >= 0) res += tr(c - (y + x + 2) + 1);
  if (c - (y + 1 + n - x) >= 0) res += tr(c - (y + 1 + n - x) + 1);
  if (c - (x + 1 + n - y) >= 0) res += tr(c - (x + 1 + n - y) + 1);
  if (c - (n - x + n - y) >= 0) res += tr(c - (n - x + n - y) + 1);
  return res;
}
int main() {
  int y, x, c;
  while (scanf("%d%d%d%d", &n, &y, &x, &c) >= 1) {
    --y, --x;
    int l = -1, r = (int)2e9 + 1;
    while (l + 1 < r) {
      int m = l + (r - l) / 2;
      if (solve(y, x, m) >= (long long)c)
        r = m;
      else
        l = m;
    }
    printf("%d\n", r);
  }
  return 0;
}

```