#include <bits/stdc++.h>
using namespace std;
long long cnt(long long n, long long x, long long y, long long t) {
  long long total = 2LL * t * t + 2LL * t + 1LL;
  long long top = 0, bot = 0, left = 0, right = 0, tl = 0, tr = 0, bl = 0,
            br = 0;
  if (n - y < t) {
    right = (t - n + y) * (t - n + y);
  }
  if (y - 1LL < t) {
    left = (t - y + 1LL) * (t - y + 1LL);
  }
  if (n - x < t) {
    bot = (t - n + x) * (t - n + x);
  }
  if (x - 1LL < t) {
    top = (t - x + 1LL) * (t - x + 1LL);
  }
  if (n - y + x - 1LL + 1LL < t) {
    tr = ((t - n + y - x + 1LL - 1LL) * (t - n + y - x + 2LL - 1LL)) / 2LL;
  }
  if (n - y + n - x + 1LL < t) {
    br = ((t - n + y - n + x - 1LL) * (t - n + y - n + x + 1LL - 1LL)) / 2LL;
  }
  if (y - 1LL + x - 1LL + 1LL < t) {
    tl = ((t - y + 1LL - x + 1LL - 1LL) * (t - y + 1LL - x + 2LL - 1LL)) / 2LL;
  }
  if (y - 1LL + n - x + 1LL < t) {
    bl =
        ((t - y + 1LL - n + x - 1LL) * (t - y + 1LL - n + x + 1LL - 1LL)) / 2LL;
  }
  total -= top + bot + left + right;
  total += tl + bl + tr + br;
  return total;
}
int main() {
  long long n, x, y, c;
  scanf(" %lld %lld %lld %lld", &n, &x, &y, &c);
  if (c <= 1) {
    printf("0\n");
    return 0;
  } else {
    long long low = 1, high = 3LL * n;
    while (1) {
      long long mid = (low + high) / 2;
      long long onboard = cnt(n, x, y, mid), onboardl = cnt(n, x, y, mid - 1);
      if (onboard >= c && onboardl < c) {
        printf("%lld\n", mid);
        return 0;
      } else if (onboard >= c) {
        high = mid;
      } else {
        low = mid;
      }
    }
  }
}

### Prompt

Create a solution in cpp for 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;
long long cnt(long long n, long long x, long long y, long long t) {
  long long total = 2LL * t * t + 2LL * t + 1LL;
  long long top = 0, bot = 0, left = 0, right = 0, tl = 0, tr = 0, bl = 0,
            br = 0;
  if (n - y < t) {
    right = (t - n + y) * (t - n + y);
  }
  if (y - 1LL < t) {
    left = (t - y + 1LL) * (t - y + 1LL);
  }
  if (n - x < t) {
    bot = (t - n + x) * (t - n + x);
  }
  if (x - 1LL < t) {
    top = (t - x + 1LL) * (t - x + 1LL);
  }
  if (n - y + x - 1LL + 1LL < t) {
    tr = ((t - n + y - x + 1LL - 1LL) * (t - n + y - x + 2LL - 1LL)) / 2LL;
  }
  if (n - y + n - x + 1LL < t) {
    br = ((t - n + y - n + x - 1LL) * (t - n + y - n + x + 1LL - 1LL)) / 2LL;
  }
  if (y - 1LL + x - 1LL + 1LL < t) {
    tl = ((t - y + 1LL - x + 1LL - 1LL) * (t - y + 1LL - x + 2LL - 1LL)) / 2LL;
  }
  if (y - 1LL + n - x + 1LL < t) {
    bl =
        ((t - y + 1LL - n + x - 1LL) * (t - y + 1LL - n + x + 1LL - 1LL)) / 2LL;
  }
  total -= top + bot + left + right;
  total += tl + bl + tr + br;
  return total;
}
int main() {
  long long n, x, y, c;
  scanf(" %lld %lld %lld %lld", &n, &x, &y, &c);
  if (c <= 1) {
    printf("0\n");
    return 0;
  } else {
    long long low = 1, high = 3LL * n;
    while (1) {
      long long mid = (low + high) / 2;
      long long onboard = cnt(n, x, y, mid), onboardl = cnt(n, x, y, mid - 1);
      if (onboard >= c && onboardl < c) {
        printf("%lld\n", mid);
        return 0;
      } else if (onboard >= c) {
        high = mid;
      } else {
        low = mid;
      }
    }
  }
}

```