#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c, u, l, r, d;
long long calc(long long a, long long b, long long c) {
  if (a == 0 || b == 0) return 0;
  if (c >= a + b) return a * b;
  if (a < b) swap(a, b);
  if (c <= b) return c * (c - 1) / 2;
  if (c <= a) return (c - 1 + c - b) * b / 2;
  return a * b - (a + b - c + 1) * (a + b - c) / 2;
}
long long co(long long a) {
  long long ret = 1ll + min(a, u) + min(a, l) + min(a, r) + min(a, d);
  return ret + calc(u, r, a) + calc(u, l, a) + calc(l, d, a) + calc(d, r, a);
}
int chk(long long a) { return (long long)c <= co(a); }
int main() {
  scanf("%lld%lld%lld%lld", &n, &x, &y, &c);
  if (c == 1) {
    putchar('0');
    return 0;
  }
  u = x - 1, l = y - 1, r = n - y, d = n - x;
  long long l = 1, r = 1000000000, mid, ans;
  while (l <= r) {
    mid = l + r >> 1;
    if (chk(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  printf("%lld", ans);
}

### Prompt

Please provide a CPP coded solution 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;
long long n, x, y, c, u, l, r, d;
long long calc(long long a, long long b, long long c) {
  if (a == 0 || b == 0) return 0;
  if (c >= a + b) return a * b;
  if (a < b) swap(a, b);
  if (c <= b) return c * (c - 1) / 2;
  if (c <= a) return (c - 1 + c - b) * b / 2;
  return a * b - (a + b - c + 1) * (a + b - c) / 2;
}
long long co(long long a) {
  long long ret = 1ll + min(a, u) + min(a, l) + min(a, r) + min(a, d);
  return ret + calc(u, r, a) + calc(u, l, a) + calc(l, d, a) + calc(d, r, a);
}
int chk(long long a) { return (long long)c <= co(a); }
int main() {
  scanf("%lld%lld%lld%lld", &n, &x, &y, &c);
  if (c == 1) {
    putchar('0');
    return 0;
  }
  u = x - 1, l = y - 1, r = n - y, d = n - x;
  long long l = 1, r = 1000000000, mid, ans;
  while (l <= r) {
    mid = l + r >> 1;
    if (chk(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  printf("%lld", ans);
}

```