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;
long long Right(int t) {
  long long tmp = x + t - n;
  if (tmp <= 0) return 0;
  return tmp * tmp;
}
long long Left(int t) {
  long long tmp = x - t - 1;
  if (tmp >= 0) return 0;
  return tmp * tmp;
}
long long Up(int t) {
  long long tmp = y + t - n;
  if (tmp <= 0) return 0;
  return tmp * tmp;
}
long long Down(int t) {
  long long tmp = y - t - 1;
  if (tmp >= 0) return 0;
  return tmp * tmp;
}
long long RightUp(int t) {
  long long ny = n - y + 1, nx = x + t - n - ny;
  if (nx >= 0) return nx * (nx + 1) / 2;
  return 0;
}
long long RightDown(int t) {
  long long nx = x + t - n - y;
  if (nx >= 0) return nx * (nx + 1) / 2;
  return 0;
}
long long LeftUp(int t) {
  long long ny = y + t - n - x;
  if (ny >= 0) return ny * (ny + 1) / 2;
  return 0;
}
long long LeftDown(int t) {
  long long tmp = t - x - y + 1;
  if (tmp >= 0) return tmp * (tmp + 1) / 2;
  return 0;
}
long long f(long long t) {
  long long res = 1 + 2 * t * (t + 1);
  res -= Right(t) + Left(t) + Up(t) + Down(t);
  res += RightUp(t) + RightDown(t) + LeftUp(t) + LeftDown(t);
  return res;
}
int main() {
  long long c;
  scanf("%lld %lld %lld %lld", &n, &x, &y, &c);
  long long l = -1, r = 2 * 1000000000;
  while (l + 1 < r) {
    long long mid = (l + r) >> 1;
    if (f(mid) >= c)
      r = mid;
    else
      l = mid;
  }
  cerr << f(37706);
  printf("%lld", r);
  return 0;
}
