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 long double EPS = 1e-14;
const long long maxn = 100500;
const long long mod = 1e9 + 7;
long long n, x, y;
long long cnt(long long X) { return X * (X + 1) / 2ll; }
long long calc(long long T) {
  long long ans = 0;
  long long i = T - x;
  i = min(i, y);
  i = max(i, 0LL);
  ans = i * x;
  ans += cnt(T - i) - cnt(max(0LL, T - y));
  return ans;
}
long long f(long long T) {
  long long tx = x, ty = y;
  long long ans = calc(T + 1);
  x = n - ty + 1, y = tx;
  ans += calc(T + 1);
  x = n - tx + 1, y = n - ty + 1;
  ans += calc(T + 1);
  x = ty, y = n - tx + 1;
  ans += calc(T + 1);
  x = tx;
  y = ty;
  return ans - min(T, n - x) - min(T, x - 1) - min(T, y - 1) - min(T, n - y) -
         3;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long c;
  cin >> n >> x >> y >> c;
  long long lo = 0, hi = 1e9;
  while (lo <= hi) {
    long long mid = (lo + hi) >> 1;
    if (f(mid) >= c)
      hi = mid - 1;
    else
      lo = mid + 1;
  }
  cout << hi + 1 << '\n';
  return 0;
}
