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>
#pragma comment(linker, "/STACK:100000000000")
bool ascending(int i, int j) { return (i < j); }
bool descending(int i, int j) { return (i > j); }
using namespace std;
int dp[5000][5000];
int A[5000];
int n, x, y, c;
bool f(unsigned long long level) {
  unsigned long long ret = 0;
  ret = (level > 1 ? level * (level - 1) * 2 : 0) + 1;
  ret -= level > x ? (level - x) * (level - x) : 0;
  ret -= level > y ? (level - y) * (level - y) : 0;
  ret -= level > n - x + 1 ? (level - n + x - 1) * (level - n + x - 1) : 0;
  ret -= level > n - y + 1 ? (level - n + y - 1) * (level - n + y - 1) : 0;
  ret += level > x + y ? (level - x - y) * (level - x - y + 1) / 2 : 0;
  ret += level > n + 1 - x + y
             ? (level - n - 1 + x - y) * (level - n + x - y) / 2
             : 0;
  ret += level > n + 1 + x - y
             ? (level - x - n - 1 + y) * (level - x - n + y) / 2
             : 0;
  ret += level > n + 1 - x + n + 1 - y
             ? (level - n - 1 + x - n - 1 + y) * (level - n - 1 + x - n + y) / 2
             : 0;
  return ret >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  int left = 0;
  int right = 2 * n + 1;
  while (left < right - 1) {
    int mid = (left + right) / 2;
    if (f(mid))
      right = mid;
    else
      left = mid;
  }
  cout << left << endl;
}
