#include <bits/stdc++.h>
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
int main() {
  int n, x, y, c;
  long long l, r, m;
  scanf("%d %d %d %d", &n, &x, &y, &c);
  x--;
  y--;
  l = -1;
  r = n * 2;
  m = (l + r) / 2;
  while (r - l > 1) {
    long long sum = m * m * 2 + m * 2 + 1;
    if (x < m) sum -= (m - x) * (m - x);
    if (y < m) sum -= (m - y) * (m - y);
    if (n - x - 1 < m) sum -= (m - n + x + 1) * (m - n + x + 1);
    if (n - y - 1 < m) sum -= (m - n + y + 1) * (m - n + y + 1);
    if (m - x - y - 1 > 0) {
      long long z = m - x - y - 1;
      sum += z * (z + 1) / 2;
    }
    if (m - x - n + y > 0) {
      long long z = m - x - n + y;
      sum += z * (z + 1) / 2;
    }
    if (m - y - n + x > 0) {
      long long z = m - y - n + x;
      sum += z * (z + 1) / 2;
    }
    if (m + x + y - n * 2 + 1 > 0) {
      long long z = m + x + y - n * 2 + 1;
      sum += z * (z + 1) / 2;
    }
    if (sum >= c) {
      r = m;
      m = (l + r) / 2;
    } else {
      l = m;
      m = (l + r) / 2;
    }
  }
  printf("%d\n", r);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
int main() {
  int n, x, y, c;
  long long l, r, m;
  scanf("%d %d %d %d", &n, &x, &y, &c);
  x--;
  y--;
  l = -1;
  r = n * 2;
  m = (l + r) / 2;
  while (r - l > 1) {
    long long sum = m * m * 2 + m * 2 + 1;
    if (x < m) sum -= (m - x) * (m - x);
    if (y < m) sum -= (m - y) * (m - y);
    if (n - x - 1 < m) sum -= (m - n + x + 1) * (m - n + x + 1);
    if (n - y - 1 < m) sum -= (m - n + y + 1) * (m - n + y + 1);
    if (m - x - y - 1 > 0) {
      long long z = m - x - y - 1;
      sum += z * (z + 1) / 2;
    }
    if (m - x - n + y > 0) {
      long long z = m - x - n + y;
      sum += z * (z + 1) / 2;
    }
    if (m - y - n + x > 0) {
      long long z = m - y - n + x;
      sum += z * (z + 1) / 2;
    }
    if (m + x + y - n * 2 + 1 > 0) {
      long long z = m + x + y - n * 2 + 1;
      sum += z * (z + 1) / 2;
    }
    if (sum >= c) {
      r = m;
      m = (l + r) / 2;
    } else {
      l = m;
      m = (l + r) / 2;
    }
  }
  printf("%d\n", r);
  return 0;
}

```