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>
const long long inf = std::numeric_limits<long long>::max();
using namespace std;
int a[1000100];
long long x, y, n, c;
long long calc(long long sec) {
  long long a = 2 * ((sec + 1) * sec) + 1;
  sec++;
  if (sec - x > 0) {
    a -= (sec - x) * (sec - x);
  }
  if (sec - y > 0) {
    a -= (sec - y) * (sec - y);
  }
  if (sec + x - n > 0) {
    a -= (sec + x - n) * (sec + x - n);
  }
  if (sec + y - n > 0) {
    a -= (sec + y - n) * (sec + y - n);
  }
  if (sec - x - y > 0) {
    a += (sec - x - y) * (sec - x - y + 1) / 2;
  }
  if (sec + x - n - y > 0) {
    a += (sec + x - n - y) * (sec + x - n - y + 1) / 2;
  }
  if (sec + x + y - n - n > 0) {
    a += (sec + x + y - n - n) * (sec + x + y - n - n + 1) / 2;
  }
  if (sec + y - n - x > 0) {
    a += (sec + y - n - x) * (sec + y - n - x + 1) / 2;
  }
  return a;
}
long long bis(long long l, long long r) {
  long long mid = (l + r) / 2;
  if (calc(mid) == c && c != (n - 1) * (n - 1)) return mid;
  if (l == r || l == r - 1) return r;
  if (calc(mid) >= c) {
    return bis(l, mid);
  } else {
    return bis(mid, r);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x >> y >> c;
  n++;
  if (c == 1) {
    cout << 0;
    return 0;
  }
  cout << bis(0, 1000000001) << endl;
  return 0;
}
