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>.
#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long sum(long long a1, long long n, long long q) {
  return (a1 + a1 + (n - 1) * q) * n / 2;
}
bool stg(long long k) {
  long long q = sum(0, k + 1, 4) + 1;
  long long rr = x + k;
  if (rr > n) {
    q -= sum(1, (rr - n), 2);
  }
  long long ll = x - k;
  if (ll < 1) {
    q -= sum(1, (1 - ll), 2);
  }
  long long uu = y + k;
  if (uu > n) {
    q -= sum(1, (uu - n), 2);
  }
  long long dd = y - k;
  if (dd < 1) {
    q -= sum(1, (1 - dd), 2);
  }
  if (uu > n) {
    long long e = (1 + (2 * (uu - n - 1)));
    if (x + ((e - 1) / 2) > n) {
      q += sum(1, ((x + ((e - 1) / 2)) - n), 1);
    }
    if (x - ((e - 1) / 2) < 1) {
      q += sum(1, (1 - (x - ((e - 1) / 2))), 1);
    }
  }
  if (dd < 1) {
    long long e = (1 + (2 * (1 - dd - 1)));
    if (x + ((e - 1) / 2) > n) {
      q += sum(1, ((x + ((e - 1) / 2)) - n), 1);
    }
    if (x - ((e - 1) / 2) < 1) {
      q += sum(1, (1 - (x - ((e - 1) / 2))), 1);
    }
  }
  return (q >= c);
}
long long byn() {
  long long l = 0, r = n + n + n;
  while ((r - l) > 3) {
    long long m = (l + r) / 2;
    if (stg(m))
      r = m;
    else
      l = m;
  }
  for (long long m = l; m <= r; ++m) {
    if (stg(m)) return m;
  }
}
int main() {
  cin >> n >> x >> y >> c;
  cout << byn() << endl;
  ;
  return 0;
}
