#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, x, y, c;
  scanf("%I64d %I64d %I64d %I64d", &n, &x, &y, &c);
  long long lo = 0;
  long long hi = 1000000000;
  vector<long long> corners(4);
  corners[0] = abs(x - 0) + abs(y - 0);
  corners[1] = abs(x - (n + 1)) + abs(y - 0);
  corners[2] = abs(x - 0) + abs(y - (n + 1));
  corners[3] = abs(x - (n + 1)) + abs(y - (n + 1));
  vector<long long> sides(4);
  sides[0] = abs(n + 1 - x);
  sides[1] = abs(0 - x);
  sides[2] = abs(n + 1 - y);
  sides[3] = abs(0 - y);
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    long long total = 0;
    total += 1 + 2LL * mid * (mid + 1);
    for (int i = 0; i < 4; i++) {
      long long prop = mid - sides[i] + 1;
      if (prop >= 1) {
        total -= prop * prop;
      }
    }
    for (int i = 0; i < 4; i++) {
      long long prop = mid - corners[i] + 1;
      if (prop >= 1) {
        total += (prop * (prop + 1)) / 2;
      }
    }
    bool possible = true;
    if (total < c) {
      possible = false;
    }
    if (!possible) {
      lo = mid + 1;
    } else {
      hi = mid;
    }
  }
  printf("%I64d\n", lo);
  return 0;
}

### Prompt

Please create a solution in cpp 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 main() {
  long long n, x, y, c;
  scanf("%I64d %I64d %I64d %I64d", &n, &x, &y, &c);
  long long lo = 0;
  long long hi = 1000000000;
  vector<long long> corners(4);
  corners[0] = abs(x - 0) + abs(y - 0);
  corners[1] = abs(x - (n + 1)) + abs(y - 0);
  corners[2] = abs(x - 0) + abs(y - (n + 1));
  corners[3] = abs(x - (n + 1)) + abs(y - (n + 1));
  vector<long long> sides(4);
  sides[0] = abs(n + 1 - x);
  sides[1] = abs(0 - x);
  sides[2] = abs(n + 1 - y);
  sides[3] = abs(0 - y);
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    long long total = 0;
    total += 1 + 2LL * mid * (mid + 1);
    for (int i = 0; i < 4; i++) {
      long long prop = mid - sides[i] + 1;
      if (prop >= 1) {
        total -= prop * prop;
      }
    }
    for (int i = 0; i < 4; i++) {
      long long prop = mid - corners[i] + 1;
      if (prop >= 1) {
        total += (prop * (prop + 1)) / 2;
      }
    }
    bool possible = true;
    if (total < c) {
      possible = false;
    }
    if (!possible) {
      lo = mid + 1;
    } else {
      hi = mid;
    }
  }
  printf("%I64d\n", lo);
  return 0;
}

```