#include <bits/stdc++.h>
unsigned long long int n, x, y, c;
void shuru() { scanf("%I64u %I64u %I64u %I64u", &n, &x, &y, &c); }
unsigned long long int f(unsigned long long int k) {
  return 2 * k * k + 2 * k + 1;
}
unsigned long long int g(unsigned long long int k) { return (k + 1) * (k + 1); }
unsigned long long int h(unsigned long long int k) {
  return (k + 1) * (k + 2) / 2;
}
unsigned long long int cnt(unsigned long long int i) {
  unsigned long long int ans = f(i);
  if (i >= x) ans -= g(i - x);
  if (i >= y) ans -= g(i - y);
  if (i >= n - x + 1) ans -= g(i - (n - x + 1));
  if (i >= n - y + 1) ans -= g(i - (n - y + 1));
  if (i >= x + y) ans += h(i - (x + y));
  if (i >= n - x + 1 + y) ans += h(i - (n - x + 1 + y));
  if (i >= x + n - y + 1) ans += h(i - (x + n - y + 1));
  if (i >= n - x + 1 + n - y + 1) ans += h(i - (n - x + 1 + n - y + 1));
  return ans;
}
void truli() {
  unsigned long long int ans = -1;
  for (unsigned long long int x = 1ll << 30; x; x /= 2)
    if (cnt(ans + x) < c) ans += x;
  printf("%I64u\n", ans + 1);
}
int main() {
  shuru();
  truli();
  return 0;
}

### Prompt

Please formulate 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>
unsigned long long int n, x, y, c;
void shuru() { scanf("%I64u %I64u %I64u %I64u", &n, &x, &y, &c); }
unsigned long long int f(unsigned long long int k) {
  return 2 * k * k + 2 * k + 1;
}
unsigned long long int g(unsigned long long int k) { return (k + 1) * (k + 1); }
unsigned long long int h(unsigned long long int k) {
  return (k + 1) * (k + 2) / 2;
}
unsigned long long int cnt(unsigned long long int i) {
  unsigned long long int ans = f(i);
  if (i >= x) ans -= g(i - x);
  if (i >= y) ans -= g(i - y);
  if (i >= n - x + 1) ans -= g(i - (n - x + 1));
  if (i >= n - y + 1) ans -= g(i - (n - y + 1));
  if (i >= x + y) ans += h(i - (x + y));
  if (i >= n - x + 1 + y) ans += h(i - (n - x + 1 + y));
  if (i >= x + n - y + 1) ans += h(i - (x + n - y + 1));
  if (i >= n - x + 1 + n - y + 1) ans += h(i - (n - x + 1 + n - y + 1));
  return ans;
}
void truli() {
  unsigned long long int ans = -1;
  for (unsigned long long int x = 1ll << 30; x; x /= 2)
    if (cnt(ans + x) < c) ans += x;
  printf("%I64u\n", ans + 1);
}
int main() {
  shuru();
  truli();
  return 0;
}

```