#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
long long n, c, x, y, res = 1, now = 1;
signed main() {
  n = read(), x = read(), y = read(), c = read();
  if (c == 1) {
    puts("0");
    return 0;
  }
  for (; c > res; now++) {
    long long L = min(now, y - 1);
    if (L == now) res--;
    res += max(0ll, L - max(now - x + 1, 1ll) + 1) +
           max(0ll, L - max(x + now - n, 1ll) + 1);
    long long R = min(now, n - y);
    if (R == now) res--;
    res += max(0ll, R - max(now - x + 1, 0ll) + 1) +
           max(0ll, R - max(x + now - n, 0ll) + 1);
  }
  printf("%lld\n", --now);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
long long n, c, x, y, res = 1, now = 1;
signed main() {
  n = read(), x = read(), y = read(), c = read();
  if (c == 1) {
    puts("0");
    return 0;
  }
  for (; c > res; now++) {
    long long L = min(now, y - 1);
    if (L == now) res--;
    res += max(0ll, L - max(now - x + 1, 1ll) + 1) +
           max(0ll, L - max(x + now - n, 1ll) + 1);
    long long R = min(now, n - y);
    if (R == now) res--;
    res += max(0ll, R - max(now - x + 1, 0ll) + 1) +
           max(0ll, R - max(x + now - n, 0ll) + 1);
  }
  printf("%lld\n", --now);
  return 0;
}

```