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 mod = 1e9 + 7;
long long mod2 = 998244353;
const long long N = 1e5 + 5;
long long calc(long long c, long long res, long long cnt) {
  long long ans = 0;
  if (c > res) {
    long long y = c - res;
    y = min(y, cnt);
    cnt -= y;
    ans = res * y;
    c = res;
  }
  ans += (cnt * (2LL * c - cnt + 1LL)) / 2LL;
  return ans;
}
long long n, x, y, c;
bool check(long long t) {
  long long temp = calc(t + 1, y, min(x, t + 1)) + calc(t, n - y, min(x, t)) +
                   calc(t, y, min(n - x, t)) +
                   calc(max(t - 1, 0LL), n - y, min(n - x, max(t - 1, 0LL)));
  return temp >= c;
}
void solve() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 4e9;
  while (l + 1 < r) {
    long long mid = l + (r - l) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  if (l < r) {
    if (check(l))
      cout << l;
    else
      cout << r;
  } else {
    if (check(r))
      cout << r;
    else
      cout << l << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
