#include <bits/stdc++.h>
using namespace std;
long long int N, X, Y, C;
inline long long int sqr(long long int X) {
  if (X <= 0) return 0LL;
  return X * X;
}
inline long long int SumFirstX(long long int X) {
  if (X <= 0) return 0LL;
  return X * (X + 1LL) / 2LL;
}
int main(void) {
  scanf("%I64d %I64d %I64d %I64d", &N, &X, &Y, &C);
  long long int L = 0LL, R = 2LL * N, Sol = 2LL * N;
  while (L <= R) {
    long long int M = (L + R) >> 1;
    long long int Cnt = 2LL * M * M + 2LL * M + 1LL;
    Cnt -= sqr(M - X + 1);
    Cnt -= sqr(M - N + X);
    Cnt -= sqr(M - Y + 1);
    Cnt -= sqr(M - N + Y);
    Cnt += SumFirstX(M - X - Y + 1);
    Cnt += SumFirstX(M - N + X - Y);
    Cnt += SumFirstX(M - N + X - N + Y - 1);
    Cnt += SumFirstX(M - X - N + Y);
    if (Cnt >= C) {
      Sol = M;
      R = M - 1;
    } else
      L = M + 1;
  }
  printf("%I64d\n", Sol);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int N, X, Y, C;
inline long long int sqr(long long int X) {
  if (X <= 0) return 0LL;
  return X * X;
}
inline long long int SumFirstX(long long int X) {
  if (X <= 0) return 0LL;
  return X * (X + 1LL) / 2LL;
}
int main(void) {
  scanf("%I64d %I64d %I64d %I64d", &N, &X, &Y, &C);
  long long int L = 0LL, R = 2LL * N, Sol = 2LL * N;
  while (L <= R) {
    long long int M = (L + R) >> 1;
    long long int Cnt = 2LL * M * M + 2LL * M + 1LL;
    Cnt -= sqr(M - X + 1);
    Cnt -= sqr(M - N + X);
    Cnt -= sqr(M - Y + 1);
    Cnt -= sqr(M - N + Y);
    Cnt += SumFirstX(M - X - Y + 1);
    Cnt += SumFirstX(M - N + X - Y);
    Cnt += SumFirstX(M - N + X - N + Y - 1);
    Cnt += SumFirstX(M - X - N + Y);
    if (Cnt >= C) {
      Sol = M;
      R = M - 1;
    } else
      L = M + 1;
  }
  printf("%I64d\n", Sol);
  return 0;
}

```