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;
const int INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int maxN = 1e5;
long long n, row, col, c;
long long sum(long long t) {
  long long max_l = col - 1, max_r = n - col;
  long long ans = 0;
  ans += 1 + min(t, max_l) + min(t, max_r);
  long long aux_t = t;
  for (int i = row - 1; i >= 1; i--) {
    aux_t--;
    if (aux_t < 0) break;
    ans += 1 + min(aux_t, max_l) + min(aux_t, max_r);
  }
  for (int i = row + 1; i <= n; i++) {
    t--;
    if (t < 0) break;
    ans += 1 + min(t, max_l) + min(t, max_r);
  }
  return ans;
}
int main() {
  scanf("%lld %lld", &n, &row), scanf("%lld %lld", &col, &c);
  long long lo = 0, hi = 1e5;
  while (lo < hi) {
    long long mi = lo + (hi - lo) / 2;
    if (sum(mi) >= c)
      hi = mi;
    else
      lo = mi + 1;
  }
  printf("%lld\n", hi);
  return 0;
}
