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;
long long X, Y;
long long C;
long long calcSide(long long l) {
  if (l < 0) return 0;
  return l * l;
}
long long calcLeft(long long t) { return calcSide(t - Y); }
long long calcRight(long long t) { return calcSide(t - (N - Y) + 1); }
long long calcTop(long long t) { return calcSide(t - X); }
long long calcBottom(long long t) { return calcSide(t - (N - X) + 1); }
long long calcCorner(long long t) {
  if (t < 0) return 0;
  return t * (t + 1) / 2;
}
long long calcLeftTop(long long t) { return calcCorner(t - X - Y - 1); }
long long calcLeftBottom(long long t) {
  return calcCorner(t - (N - 1 - X) - Y - 1);
}
long long calcRightTop(long long t) {
  return calcCorner(t - X - (N - 1 - Y) - 1);
}
long long calcRightBottom(long long t) {
  return calcCorner(t - (N - 1 - X) - (N - 1 - Y) - 1);
}
long long calc(long long t) {
  long long res = t * t + (t + 1) * (t + 1);
  res -= calcLeft(t);
  res -= calcRight(t);
  res -= calcTop(t);
  res -= calcBottom(t);
  res += calcLeftTop(t);
  res += calcLeftBottom(t);
  res += calcRightTop(t);
  res += calcRightBottom(t);
  return res;
}
int main() {
  cin >> N >> X >> Y >> C;
  X--;
  Y--;
  long long a = 0;
  long long b = 2000000000;
  while (b - a > 2) {
    long long c = (a + b) / 2;
    long long q = calc(c);
    if (q >= C)
      b = c;
    else
      a = c;
  }
  long long res = b;
  for (long long i = a; i < b; i++)
    if (calc(i) >= C) {
      res = i;
      break;
    }
  cout << res << "\n";
  return 0;
}
