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 std::abs;
using std::bitset;
using std::cerr;
using std::cin;
using std::copy;
using std::cout;
using std::deque;
using std::endl;
using std::fill;
using std::fixed;
using std::greater;
using std::istream;
using std::lower_bound;
using std::make_pair;
using std::make_tuple;
using std::map;
using std::max;
using std::max_element;
using std::min;
using std::min_element;
using std::multiset;
using std::next_permutation;
using std::nth_element;
using std::ostream;
using std::pair;
using std::priority_queue;
using std::queue;
using std::reverse;
using std::rotate;
using std::set;
using std::sort;
using std::string;
using std::swap;
using std::tie;
using std::tuple;
using std::unique;
using std::unordered_map;
using std::unordered_set;
using std::upper_bound;
using std::vector;
int const INF = (int)1e9;
long long const INFL = (long long)1e18;
long double const PI = 3.1415926535897932384626433832795028;
long long sumTo(long long x) { return x * (x + 1) / 2; }
long long go(long long x, long long y, long long t) {
  --t;
  if (t <= 0) return 0;
  if (x > y) swap(x, y);
  if (t <= x) {
    return sumTo(t);
  }
  if (t <= y) {
    return sumTo(t) - sumTo(t - x);
  }
  if (t >= x + y - 1) {
    return x * y;
  }
  return x * y - sumTo(x + y - 1 - t);
}
void solve() {
  int n, x, y;
  long long c;
  cin >> n >> x >> y >> c;
  int l = -1;
  int r = n + n;
  while (r - l > 1) {
    int mid = l + (r - l) / 2;
    long long cnt = 1 + (long long)min(x - 1, mid) +
                    (long long)min(n - x, mid) + (long long)min(y - 1, mid) +
                    (long long)min(n - y, mid);
    cnt += go(x - 1, y - 1, mid) + go(x - 1, n - y, mid) +
           go(n - x, y - 1, mid) + go(n - x, n - y, mid);
    if (cnt >= c) {
      r = mid;
    } else {
      l = mid;
    }
  }
  cout << r << endl;
}
int main() {
  cout.precision(15);
  cout << fixed;
  cerr.precision(6);
  cerr << fixed;
  srand(1510);
  solve();
}
