#include <bits/stdc++.h>
using namespace std;
int n, x, y, c;
bool check(int i, int j) {
  if (i < 1 || j < 1) return false;
  if (i > n || j > n) return false;
  return true;
}
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int main() {
  scanf("%d", &n);
  scanf("%d", &x);
  scanf("%d", &y);
  scanf("%d", &c);
  int sec = 0, cnt = 1;
  while (cnt < c) {
    sec++;
    int nx, ny, nx2, ny2;
    ny = max(1, y - sec), nx = x + y - sec - ny;
    nx2 = max(1, x - sec), ny2 = x + y - sec - nx2;
    if (ny2 >= ny) cnt += (ny2 - ny + 1);
    ny = min(n, y + sec), nx = x + y + sec - ny;
    nx2 = min(n, x + sec), ny2 = x + y + sec - nx2;
    if (ny >= ny2) cnt += (ny - ny2 + 1);
    ny = max(1, y - sec), nx = x - y + sec + ny;
    nx2 = min(n, x + sec), ny2 = nx2 - (x - y + sec);
    if (ny2 >= ny) cnt += (ny2 - ny + 1);
    ny = min(n, y + sec), nx = x - y - sec + ny;
    nx2 = max(1, x - sec), ny2 = nx2 - (x - y - sec);
    if (ny >= ny2) cnt += (ny - ny2 + 1);
    for (int i = 0; i < 4; i++) {
      nx = x + dx[i] * sec, ny = y + dy[i] * sec;
      cnt -= check(nx, ny);
    }
  }
  cout << sec << endl;
  return 0;
}

### Prompt

Your task is to create 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;
int n, x, y, c;
bool check(int i, int j) {
  if (i < 1 || j < 1) return false;
  if (i > n || j > n) return false;
  return true;
}
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int main() {
  scanf("%d", &n);
  scanf("%d", &x);
  scanf("%d", &y);
  scanf("%d", &c);
  int sec = 0, cnt = 1;
  while (cnt < c) {
    sec++;
    int nx, ny, nx2, ny2;
    ny = max(1, y - sec), nx = x + y - sec - ny;
    nx2 = max(1, x - sec), ny2 = x + y - sec - nx2;
    if (ny2 >= ny) cnt += (ny2 - ny + 1);
    ny = min(n, y + sec), nx = x + y + sec - ny;
    nx2 = min(n, x + sec), ny2 = x + y + sec - nx2;
    if (ny >= ny2) cnt += (ny - ny2 + 1);
    ny = max(1, y - sec), nx = x - y + sec + ny;
    nx2 = min(n, x + sec), ny2 = nx2 - (x - y + sec);
    if (ny2 >= ny) cnt += (ny2 - ny + 1);
    ny = min(n, y + sec), nx = x - y - sec + ny;
    nx2 = max(1, x - sec), ny2 = nx2 - (x - y - sec);
    if (ny >= ny2) cnt += (ny - ny2 + 1);
    for (int i = 0; i < 4; i++) {
      nx = x + dx[i] * sec, ny = y + dy[i] * sec;
      cnt -= check(nx, ny);
    }
  }
  cout << sec << endl;
  return 0;
}

```