#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long mx(long long a, long long b) { return a > b ? a : b; }
long long mn(long long a, long long b) { return a > b ? b : a; }
long long getv(long long t) {
  long long top, botton, ul, ur, ml, mr, dl, dr;
  if (y - t < 1) {
    ul = mx(1, x - abs(y - t) - 1);
    ur = mn(n, x + abs(y - t) + 1);
    top = 1;
  } else {
    ul = ur = x;
    top = y - t;
  }
  if (y + t > n) {
    dl = mx(1, x - (y + t - n));
    dr = mn(n, x + (y + t - n));
    botton = n;
  } else {
    dl = dr = x;
    botton = y + t;
  }
  ml = mx(1, x - t);
  mr = mn(n, x + t);
  long long u = x - ul + 1, m = x - ml + 1, d = x - dl + 1;
  long long tt = m - u, mt = y - top + 1 - (m - u);
  long long lu = tt * (u + u + tt - 1) / 2 + mt * m;
  tt = m - d, mt = botton - y - (m - d);
  long long ld = tt * (d + d + tt - 1) / 2 + mt * m;
  u = ur - x, m = mr - x, d = dr - x;
  tt = m - u, mt = y - top + 1 - (m - u);
  long long ru = tt * (u + u + tt - 1) / 2 + mt * m;
  tt = m - d, mt = botton - y - (m - d);
  long long rd = tt * (d + d + tt - 1) / 2 + mt * m;
  return lu + ld + ru + rd;
}
int main() {
  while (cin >> n >> y >> x >> c) {
    long long l = 0, r = n * n, mid;
    while (l < r) {
      if (getv(l) >= c) break;
      mid = (l + r) >> 1;
      if (getv(mid) < c)
        l = mid + 1;
      else
        r = mid;
    }
    cout << l << endl;
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
long long n, x, y, c;
long long mx(long long a, long long b) { return a > b ? a : b; }
long long mn(long long a, long long b) { return a > b ? b : a; }
long long getv(long long t) {
  long long top, botton, ul, ur, ml, mr, dl, dr;
  if (y - t < 1) {
    ul = mx(1, x - abs(y - t) - 1);
    ur = mn(n, x + abs(y - t) + 1);
    top = 1;
  } else {
    ul = ur = x;
    top = y - t;
  }
  if (y + t > n) {
    dl = mx(1, x - (y + t - n));
    dr = mn(n, x + (y + t - n));
    botton = n;
  } else {
    dl = dr = x;
    botton = y + t;
  }
  ml = mx(1, x - t);
  mr = mn(n, x + t);
  long long u = x - ul + 1, m = x - ml + 1, d = x - dl + 1;
  long long tt = m - u, mt = y - top + 1 - (m - u);
  long long lu = tt * (u + u + tt - 1) / 2 + mt * m;
  tt = m - d, mt = botton - y - (m - d);
  long long ld = tt * (d + d + tt - 1) / 2 + mt * m;
  u = ur - x, m = mr - x, d = dr - x;
  tt = m - u, mt = y - top + 1 - (m - u);
  long long ru = tt * (u + u + tt - 1) / 2 + mt * m;
  tt = m - d, mt = botton - y - (m - d);
  long long rd = tt * (d + d + tt - 1) / 2 + mt * m;
  return lu + ld + ru + rd;
}
int main() {
  while (cin >> n >> y >> x >> c) {
    long long l = 0, r = n * n, mid;
    while (l < r) {
      if (getv(l) >= c) break;
      mid = (l + r) >> 1;
      if (getv(mid) < c)
        l = mid + 1;
      else
        r = mid;
    }
    cout << l << endl;
  }
  return 0;
}

```