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 N, X, Y, C;
long long D[5];
void read(void);
void soups_on(void);
long long yee_haw(long long);
int main() {
  read(), soups_on();
  return 0;
}
void read(void) { cin >> N >> X >> Y >> C, --X, --Y; }
void soups_on(void) {
  long long low = 0, up = 0, mid;
  D[0] = X, D[1] = Y, D[2] = N - 1 - X, D[3] = N - 1 - Y, D[4] = D[0];
  for (int i = 0; i < 4; ++i) up = max(up, D[i] + D[i + 1]);
  while (low < up) {
    mid = (low + up) / 2;
    if (yee_haw(mid) < C)
      low = mid + 1;
    else
      up = mid;
  }
  cout << low << "\n";
}
long long yee_haw(long long b) {
  long long res = 1;
  for (int i = 0; i < 4; ++i) {
    if (D[i] + D[i + 1] <= b)
      res += (D[i] + 1) * (D[i + 1] + 1);
    else {
      res += (b + 1) * (b + 2) / 2;
      if (D[i] < b) res -= (b - D[i]) * (b - D[i] + 1) / 2;
      if (D[i + 1] < b) res -= (b - D[i + 1]) * (b - D[i + 1] + 1) / 2;
    }
    res -= min(D[i], b) + 1;
  }
  return res;
}
