#include <bits/stdc++.h>
using namespace std;
int n, x, y;
long long c;
long long deal(int x, int y, int t) {
  int p = x + y - t - 1;
  if (p <= 0) return 0;
  long long ans = (long long)p * (p + 1) / 2;
  return ans;
}
int OK(int t) {
  long long s = 2 * (long long)t * t + 2 * t + 1;
  long long temp, p[4];
  if (t + y > n) {
    temp = t + y - n;
    s -= temp * temp;
    p[0] = temp;
  } else
    p[0] = -1;
  if (x - t < 1) {
    temp = t - x + 1;
    s -= temp * temp;
    p[1] = temp;
  } else
    p[1] = -1;
  if (y - t < 1) {
    temp = t + 1 - y;
    s -= temp * temp;
    p[2] = temp;
  } else
    p[2] = -1;
  if (t + x > n) {
    temp = t + x - n;
    s -= temp * temp;
    p[3] = temp;
  } else
    p[3] = -1;
  if (p[0] != -1 && p[1] != -1) s += deal(p[0], p[1], t);
  if (p[2] != -1 && p[1] != -1) s += deal(p[1], p[2], t);
  if (p[2] != -1 && p[3] != -1) s += deal(p[2], p[3], t);
  if (p[0] != -1 && p[3] != -1) s += deal(p[0], p[3], t);
  return s >= c;
}
int main() {
  while (scanf("%d%d%d%I64d", &n, &x, &y, &c) != -1) {
    int low = 1, high = n + n - 1, mid;
    while (low <= high) {
      mid = (low + high) >> 1;
      if (OK(mid))
        high = mid - 1;
      else
        low = mid + 1;
    }
    int ans;
    if (OK(high))
      ans = high;
    else
      ans = low;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
int n, x, y;
long long c;
long long deal(int x, int y, int t) {
  int p = x + y - t - 1;
  if (p <= 0) return 0;
  long long ans = (long long)p * (p + 1) / 2;
  return ans;
}
int OK(int t) {
  long long s = 2 * (long long)t * t + 2 * t + 1;
  long long temp, p[4];
  if (t + y > n) {
    temp = t + y - n;
    s -= temp * temp;
    p[0] = temp;
  } else
    p[0] = -1;
  if (x - t < 1) {
    temp = t - x + 1;
    s -= temp * temp;
    p[1] = temp;
  } else
    p[1] = -1;
  if (y - t < 1) {
    temp = t + 1 - y;
    s -= temp * temp;
    p[2] = temp;
  } else
    p[2] = -1;
  if (t + x > n) {
    temp = t + x - n;
    s -= temp * temp;
    p[3] = temp;
  } else
    p[3] = -1;
  if (p[0] != -1 && p[1] != -1) s += deal(p[0], p[1], t);
  if (p[2] != -1 && p[1] != -1) s += deal(p[1], p[2], t);
  if (p[2] != -1 && p[3] != -1) s += deal(p[2], p[3], t);
  if (p[0] != -1 && p[3] != -1) s += deal(p[0], p[3], t);
  return s >= c;
}
int main() {
  while (scanf("%d%d%d%I64d", &n, &x, &y, &c) != -1) {
    int low = 1, high = n + n - 1, mid;
    while (low <= high) {
      mid = (low + high) >> 1;
      if (OK(mid))
        high = mid - 1;
      else
        low = mid + 1;
    }
    int ans;
    if (OK(high))
      ans = high;
    else
      ans = low;
    printf("%d\n", ans);
  }
  return 0;
}

```