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;
bool check(long long T) {
  long long ans = T * T + (T + 1) * (T + 1);
  if (x + T >= n) ans -= (T - (n - 1 - x)) * (T - (n - 1 - x));
  if (x - T < 0) ans -= (T - x) * (T - x);
  if (y + T >= n) ans -= (T - (n - 1 - y)) * (T - (n - 1 - y));
  if (y - T < 0) ans -= (T - y) * (T - y);
  if (n - x + n - y <= T) {
    long long t = T - (n - x + n - y) + 1;
    ans += t * (t + 1) / 2;
  }
  if (n - x + y + 1 <= T) {
    long long t = T - (n - x + y + 1) + 1;
    ans += t * (t + 1) / 2;
  }
  if (x + 1 + n - y <= T) {
    long long t = T - (x + 1 + n - y) + 1;
    ans += t * (t + 1) / 2;
  }
  if (x + 1 + y + 1 <= T) {
    long long t = T - (x + 1 + y + 1) + 1;
    ans += t * (t + 1) / 2;
  }
  return ans >= c;
}
void solve() {
  cin >> n >> x >> y >> c;
  x--, y--;
  long long lo = 0, hi = 1e9;
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    if (check(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  cout << lo << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
