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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
long long n, x, y, c;
long long calcC(long long dis1, long long dis2, long long m) {
  long long hm = max(0LL, m - (dis1 + dis2));
  return (hm * hm + hm) / 2;
}
long long calcB(long long dis, long long m) {
  long long hm = max(0LL, m - dis);
  return hm * hm;
}
bool solve(long long m) {
  long long ar = 0, dis;
  ar += ((m * m) - m) * 2 + 1;
  ar -= calcB(x, m);
  ar -= calcB(n - x + 1, m);
  ar -= calcB(y, m);
  ar -= calcB(n - y + 1, m);
  ar += calcC(x, y, m);
  ar += calcC(n - x + 1, y, m);
  ar += calcC(x, n - y + 1, m);
  ar += calcC(n - x + 1, n - y + 1, m);
  if (ar >= c)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> x >> y >> c;
  long long l = 1, r = 2 * n;
  while (l < r) {
    long long m = (l + r) / 2;
    if (solve(m))
      r = m;
    else
      l = m + 1;
  }
  cout << l - 1 << '\n';
  exit(0);
}
