#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using std::bitset;
using namespace std;
long long n, x, y, c;
long long ans;
long long judge(long long t) {
  long long i, j, tem, sum = 0, a, b;
  long long l, r, u, d;
  l = max(0ll, t - y + 1);
  r = max(0ll, t - n + y);
  u = max(0ll, t - x + 1);
  d = max(0ll, t - n + x);
  sum = 2 * t * t + 2 * t + 1 - l * l - r * r - u * u - d * d;
  a = max(0ll, u - n + y - 1);
  b = max(0ll, r - x + 1 - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  a = max(0ll, u - y + 1 - 1);
  b = max(0ll, l - x + 1 - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  a = max(0ll, d - y + 1 - 1);
  b = max(0ll, l - n + x - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  a = max(0ll, d - n + y - 1);
  b = max(0ll, r - n + x - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  return sum >= c;
}
int main() {
  long long i, j, tem;
  long long low, high, mid;
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  low = 0;
  high = 2 * n;
  while (low <= high) {
    mid = (low + high) / 2;
    if (judge(mid)) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using std::bitset;
using namespace std;
long long n, x, y, c;
long long ans;
long long judge(long long t) {
  long long i, j, tem, sum = 0, a, b;
  long long l, r, u, d;
  l = max(0ll, t - y + 1);
  r = max(0ll, t - n + y);
  u = max(0ll, t - x + 1);
  d = max(0ll, t - n + x);
  sum = 2 * t * t + 2 * t + 1 - l * l - r * r - u * u - d * d;
  a = max(0ll, u - n + y - 1);
  b = max(0ll, r - x + 1 - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  a = max(0ll, u - y + 1 - 1);
  b = max(0ll, l - x + 1 - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  a = max(0ll, d - y + 1 - 1);
  b = max(0ll, l - n + x - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  a = max(0ll, d - n + y - 1);
  b = max(0ll, r - n + x - 1);
  tem = min(a, b);
  sum += (1 + tem) * tem / 2;
  return sum >= c;
}
int main() {
  long long i, j, tem;
  long long low, high, mid;
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  low = 0;
  high = 2 * n;
  while (low <= high) {
    mid = (low + high) / 2;
    if (judge(mid)) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  printf("%I64d\n", ans);
  return 0;
}

```