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;
int dis[4], size, x, y;
long long num;
long long from_one(long long n) { return (n + 1) * n / 2; }
long long cal(long long t) {
  long long ret = 2 * t * t + 2 * t + 1;
  for (int i = 0; i < 4; i++) {
    if (t <= dis[i]) continue;
    long long res = (t - dis[i]) * (t - dis[i]);
    ret -= res;
  }
  for (int i = 0; i < 4; i++) {
    if (t - (dis[i] + dis[(i + 1) % 4]) - 1 <= 0) continue;
    long long res = from_one(t - (dis[i] + dis[(i + 1) % 4]) - 1);
    ret += res;
  }
  return ret;
}
int main() {
  cin >> size >> x >> y >> num;
  if (num == 1) {
    puts("0");
    return 0;
  }
  dis[0] = size - x;
  dis[1] = size - y;
  dis[2] = x - 1;
  dis[3] = y - 1;
  long long b = 0, e = 2 * (size - 1);
  while (b + 1 < e) {
    long long mid = (b + e) / 2;
    if (cal(mid) < num) {
      b = mid;
    } else {
      e = mid;
    }
  }
  if (cal(e) >= num)
    cout << e << endl;
  else
    cout << b << endl;
}
