#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000009;
const long long INFL = 1000000000LL * 1000000000LL + 1000LL;
const int INF = 1000000000 + 1000;
long long ile(int n, int x, int y, int czas) {
  int gora = x + czas;
  int dol = x - czas;
  gora = min(n, gora);
  dol = max(1, dol);
  long long res = gora - dol + 1;
  for (int i = 1; i <= czas; i++) {
    gora = x + czas - i;
    dol = x - czas + i;
    gora = min(n, gora);
    dol = max(1, dol);
    int lewo = y - i;
    int prawo = y + i;
    int wsp = 0;
    if (lewo > 0) {
      wsp++;
    }
    if (prawo <= n) {
      wsp++;
    }
    if (wsp == 0) {
      break;
    }
    res += (gora - dol + 1) * wsp;
  }
  return res;
}
int n, x, y, c;
int main() {
  scanf("%d %d %d %d", &n, &x, &y, &c);
  int l = 0, r = 100000, m;
  while (l < r) {
    m = (l + r) / 2;
    if (ile(n, x, y, m) >= c) {
      r = m;
    } else {
      l = m + 1;
    }
  }
  printf("%d\n", l);
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
using namespace std;
const long long MOD = 1000000009;
const long long INFL = 1000000000LL * 1000000000LL + 1000LL;
const int INF = 1000000000 + 1000;
long long ile(int n, int x, int y, int czas) {
  int gora = x + czas;
  int dol = x - czas;
  gora = min(n, gora);
  dol = max(1, dol);
  long long res = gora - dol + 1;
  for (int i = 1; i <= czas; i++) {
    gora = x + czas - i;
    dol = x - czas + i;
    gora = min(n, gora);
    dol = max(1, dol);
    int lewo = y - i;
    int prawo = y + i;
    int wsp = 0;
    if (lewo > 0) {
      wsp++;
    }
    if (prawo <= n) {
      wsp++;
    }
    if (wsp == 0) {
      break;
    }
    res += (gora - dol + 1) * wsp;
  }
  return res;
}
int n, x, y, c;
int main() {
  scanf("%d %d %d %d", &n, &x, &y, &c);
  int l = 0, r = 100000, m;
  while (l < r) {
    m = (l + r) / 2;
    if (ile(n, x, y, m) >= c) {
      r = m;
    } else {
      l = m + 1;
    }
  }
  printf("%d\n", l);
}

```