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 pintados(long long segundos) {
  long long pos, aArriba = 0, aAbajo = 0, contador = 0;
  bool arriba = true, abajo = true;
  for (pos = x - segundos; pos < 1; ++pos) {
    if (y + aAbajo == n) abajo = false;
    if (y - aArriba == 1) arriba = false;
    if (abajo) ++aAbajo;
    if (arriba) ++aArriba;
  }
  for (pos; pos <= x and pos <= n; ++pos) {
    contador += aAbajo + aArriba + 1;
    if (contador < 0) return INT_MAX;
    if (y + aAbajo == n) abajo = false;
    if (y - aArriba == 1) arriba = false;
    if (abajo) ++aAbajo;
    if (arriba) ++aArriba;
  }
  aArriba = aAbajo = 0;
  arriba = abajo = true;
  for (pos = x + segundos; pos > n; --pos) {
    if (y + aAbajo == n) abajo = false;
    if (y - aArriba == 1) arriba = false;
    if (abajo) ++aAbajo;
    if (arriba) ++aArriba;
  }
  for (pos; pos > x; --pos) {
    contador += aAbajo + aArriba + 1;
    if (contador < 0) return INT_MAX;
    if (y + aAbajo == n) abajo = false;
    if (y - aArriba == 1) arriba = false;
    if (abajo) ++aAbajo;
    if (arriba) ++aArriba;
  }
  return contador;
}
long long buscar() {
  long long izq = 0, der = min((long long)2e5, n * 2), centro;
  while (izq <= der) {
    centro = (der - izq) / 2 + izq;
    long long segundos = pintados(centro), ant = pintados(centro - 1);
    if (segundos >= c and ant < c) return centro;
    if (segundos < c)
      izq = centro + 1;
    else
      der = centro - 1;
  }
}
int main() {
  cin >> n >> x >> y >> c;
  cout << buscar() << '\n';
}
