#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
#pragma comment(linker, "/STACK:256000000")
const int D = 10000;
double solve(long long a, long long b, long long c) {
  if (a == 0) {
    return 1. * -c / b;
  }
  long long d = b * b - 4 * a * c;
  return (-b + sqrt((d)*1.)) / (2 * a);
}
int main() {
  int n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int a[5];
  a[0] = x - 1;
  a[1] = y - 1;
  a[2] = n - x;
  a[3] = n - y;
  int b[5];
  a[4] = b[4] = 1000000000;
  for (int i = 0; i < 4; i++) b[i] = a[i] + a[(i + 1) % 4];
  sort(a, a + 4);
  sort(b, b + 4);
  int cura = 0, curb = 0;
  long long s = 1;
  long long da = 4;
  long long sp = 4;
  long long t = 0;
  long long nexts = s;
  long long lasta = da;
  long long lastsp = sp;
  long long p = 0;
  while (nexts < c) {
    t += p;
    s = nexts;
    lasta = da;
    lastsp = sp;
    p = ((a[cura]) < (b[curb]) ? (a[cura]) : (b[curb])) - t;
    nexts += p * (2 * sp + da * (p - 1)) / 2;
    sp += da * p;
    if (a[cura] < b[curb]) {
      cura++;
      da -= 2;
      sp -= 1;
    } else {
      da += 1;
      curb++;
    }
  }
  double m = solve(lasta, 2 * lastsp - lasta, 2 * (s - c)) - 1E-5;
  int o = t + (int)(m + 0.9999999);
  cout << o << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>.

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
#pragma comment(linker, "/STACK:256000000")
const int D = 10000;
double solve(long long a, long long b, long long c) {
  if (a == 0) {
    return 1. * -c / b;
  }
  long long d = b * b - 4 * a * c;
  return (-b + sqrt((d)*1.)) / (2 * a);
}
int main() {
  int n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int a[5];
  a[0] = x - 1;
  a[1] = y - 1;
  a[2] = n - x;
  a[3] = n - y;
  int b[5];
  a[4] = b[4] = 1000000000;
  for (int i = 0; i < 4; i++) b[i] = a[i] + a[(i + 1) % 4];
  sort(a, a + 4);
  sort(b, b + 4);
  int cura = 0, curb = 0;
  long long s = 1;
  long long da = 4;
  long long sp = 4;
  long long t = 0;
  long long nexts = s;
  long long lasta = da;
  long long lastsp = sp;
  long long p = 0;
  while (nexts < c) {
    t += p;
    s = nexts;
    lasta = da;
    lastsp = sp;
    p = ((a[cura]) < (b[curb]) ? (a[cura]) : (b[curb])) - t;
    nexts += p * (2 * sp + da * (p - 1)) / 2;
    sp += da * p;
    if (a[cura] < b[curb]) {
      cura++;
      da -= 2;
      sp -= 1;
    } else {
      da += 1;
      curb++;
    }
  }
  double m = solve(lasta, 2 * lastsp - lasta, 2 * (s - c)) - 1E-5;
  int o = t + (int)(m + 0.9999999);
  cout << o << endl;
  return 0;
}

```