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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
long long int n, x, y;
long long int c;
long long int get1(long long int a) {
  if (a <= 0) return 0;
  return a * a;
}
long long int get2(long long int a) {
  if (a <= 0) return 0;
  return a * (a + 1) / 2;
}
long long int check(long long int md) {
  long long int area = 2 * (long long int)(md + 1) * md + 1;
  area -= get1(x + md - n + 1) + get1(y + md - n + 1) + get1(-(x - md)) +
          get1(-(y - md));
  area += get2(x + md - y - n) + get2(x + md - n + 1 - (n - y)) +
          get2(-(x - md) - (n - y)) + get2(-(x - md) - y - 1);
  return area;
}
int main() {
  cin >> n >> x >> y >> c;
  --x;
  --y;
  long long int lb = -1, ub = 1e9 + 5;
  while (ub - lb > 1) {
    long long int md = (lb + ub) >> 1;
    if (check(md) >= c)
      ub = md;
    else
      lb = md;
  }
  cout << ub << endl;
  return 0;
}
