#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x003fffff;
const int inf = -0x003fffff;
const double eps = 1e-6;
const double EPS = eps;
inline bool sless(double a, double b) { return a < b; }
inline bool sbigger(double a, double b) { return a > b; }
template <class _Template, class _Compare, class _Function>
inline _Template sanfen(_Template l, _Template r, _Compare comp, _Function f) {
  _Template mid, mmid, s1, s2;
  while (r - l > EPS) {
    mid = (l + r) / 2;
    mmid = (mid + r) / 2;
    s1 = f(mid);
    s2 = f(mmid);
    (comp(s1, s2)) ? (r = mmid) : (l = mid);
  }
  return comp(s1, s2) ? l : r;
}
void L() {
  double s, a, b, c;
  while (cin >> s >> a >> b >> c) {
    if (a + b + c < eps) {
      printf("%.12f 0.00 0.00\n", s);
      continue;
    }
    double sum = a + b + c;
    printf("%.12f %.12f %.12f\n", s * a / sum, s * b / sum, s * c / sum);
  }
}
long long n, x, y, c;
bool check(long long t) {
  long long temp[] = {(x + 0) + (y + 0),     (x + 0) + (y - y),
                      (x + 0) + (n + 1 - y), (x - x) + (y + 0),
                      (x - x) + (n + 1 - y), (n + 1 - x) + (y + 0),
                      (n + 1 - x) + (y - y), (n + 1 - x) + (n + 1 - y)};
  int op[] = {1, -1, 1, -1, -1, 1, -1, 1};
  long long sum = 0;
  for (int i = 0; i < 8; i++) {
    long long dis = t - temp[i] + 1;
    if (dis > 0)
      sum += op[i] * ((~op[i]) ? (dis * (dis + 1) / 2) : (dis * dis));
  }
  sum += 2 * t * t + 2 * t + 1;
  return (sum >= c);
}
void A() {
  while (cin >> n >> x >> y >> c) {
    long long l = 0, r = n + n;
    while (l < r) {
      long long mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid + 1;
    }
    cout << l << endl;
  }
}
int main() {
  A();
  return 0;
}

### 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x003fffff;
const int inf = -0x003fffff;
const double eps = 1e-6;
const double EPS = eps;
inline bool sless(double a, double b) { return a < b; }
inline bool sbigger(double a, double b) { return a > b; }
template <class _Template, class _Compare, class _Function>
inline _Template sanfen(_Template l, _Template r, _Compare comp, _Function f) {
  _Template mid, mmid, s1, s2;
  while (r - l > EPS) {
    mid = (l + r) / 2;
    mmid = (mid + r) / 2;
    s1 = f(mid);
    s2 = f(mmid);
    (comp(s1, s2)) ? (r = mmid) : (l = mid);
  }
  return comp(s1, s2) ? l : r;
}
void L() {
  double s, a, b, c;
  while (cin >> s >> a >> b >> c) {
    if (a + b + c < eps) {
      printf("%.12f 0.00 0.00\n", s);
      continue;
    }
    double sum = a + b + c;
    printf("%.12f %.12f %.12f\n", s * a / sum, s * b / sum, s * c / sum);
  }
}
long long n, x, y, c;
bool check(long long t) {
  long long temp[] = {(x + 0) + (y + 0),     (x + 0) + (y - y),
                      (x + 0) + (n + 1 - y), (x - x) + (y + 0),
                      (x - x) + (n + 1 - y), (n + 1 - x) + (y + 0),
                      (n + 1 - x) + (y - y), (n + 1 - x) + (n + 1 - y)};
  int op[] = {1, -1, 1, -1, -1, 1, -1, 1};
  long long sum = 0;
  for (int i = 0; i < 8; i++) {
    long long dis = t - temp[i] + 1;
    if (dis > 0)
      sum += op[i] * ((~op[i]) ? (dis * (dis + 1) / 2) : (dis * dis));
  }
  sum += 2 * t * t + 2 * t + 1;
  return (sum >= c);
}
void A() {
  while (cin >> n >> x >> y >> c) {
    long long l = 0, r = n + n;
    while (l < r) {
      long long mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid + 1;
    }
    cout << l << endl;
  }
}
int main() {
  A();
  return 0;
}

```