#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long ans = 1, v1 = -1, v2 = -1, g1 = -1, g2 = -1, vg1 = -1, vg2 = -1,
            vg3 = -1, vg4 = -1, r = 0;
  int tv1 = 0, tv2 = 0, tg1 = 0, tg2 = 0;
  int i = 0;
  int tmp;
  for (; ans < c; i++) {
    ans += r;
    r += 4;
    if (v1 != -1) v1 += 2;
    if (v2 != -1) v2 += 2;
    if (g1 != -1) g1 += 2;
    if (g2 != -1) g2 += 2;
    if (vg1 != -1) vg1++;
    if (vg2 != -1) vg2++;
    if (vg3 != -1) vg3++;
    if (vg4 != -1) vg4++;
    if (v1 == -1 && (y + i) > n) {
      v1 = 1;
      tv1 = i;
    }
    if (g1 == -1 && (x + i) > n) {
      g1 = 1;
      tg1 = i;
    }
    if (g2 == -1 && (x - i) < 1) {
      g2 = 1;
      tg2 = i;
    }
    if (v2 == -1 && (y - i) < 1) {
      v2 = 1;
      tv2 = i;
    }
    if (vg1 == -1 && tv1 != 0 && tg1 != 0 && (tv1 + tg1) <= i) vg1 = 1;
    if (vg2 == -1 && tv1 != 0 && tg2 != 0 && (tv1 + tg2) <= i) vg2 = 1;
    if (vg3 == -1 && tv2 != 0 && tg1 != 0 && (tv2 + tg1) <= i) vg3 = 1;
    if (vg4 == -1 && tv2 != 0 && tg2 != 0 && (tv2 + tg2) <= i) vg4 = 1;
    if (v1 != -1) ans -= v1;
    if (v2 != -1) ans -= v2;
    if (g1 != -1) ans -= g1;
    if (g2 != -1) ans -= g2;
    if (vg1 != -1) ans += vg1;
    if (vg2 != -1) ans += vg2;
    if (vg3 != -1) ans += vg3;
    if (vg4 != -1) ans += vg4;
    if (ans >= c) break;
  }
  cout << i;
  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;
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long ans = 1, v1 = -1, v2 = -1, g1 = -1, g2 = -1, vg1 = -1, vg2 = -1,
            vg3 = -1, vg4 = -1, r = 0;
  int tv1 = 0, tv2 = 0, tg1 = 0, tg2 = 0;
  int i = 0;
  int tmp;
  for (; ans < c; i++) {
    ans += r;
    r += 4;
    if (v1 != -1) v1 += 2;
    if (v2 != -1) v2 += 2;
    if (g1 != -1) g1 += 2;
    if (g2 != -1) g2 += 2;
    if (vg1 != -1) vg1++;
    if (vg2 != -1) vg2++;
    if (vg3 != -1) vg3++;
    if (vg4 != -1) vg4++;
    if (v1 == -1 && (y + i) > n) {
      v1 = 1;
      tv1 = i;
    }
    if (g1 == -1 && (x + i) > n) {
      g1 = 1;
      tg1 = i;
    }
    if (g2 == -1 && (x - i) < 1) {
      g2 = 1;
      tg2 = i;
    }
    if (v2 == -1 && (y - i) < 1) {
      v2 = 1;
      tv2 = i;
    }
    if (vg1 == -1 && tv1 != 0 && tg1 != 0 && (tv1 + tg1) <= i) vg1 = 1;
    if (vg2 == -1 && tv1 != 0 && tg2 != 0 && (tv1 + tg2) <= i) vg2 = 1;
    if (vg3 == -1 && tv2 != 0 && tg1 != 0 && (tv2 + tg1) <= i) vg3 = 1;
    if (vg4 == -1 && tv2 != 0 && tg2 != 0 && (tv2 + tg2) <= i) vg4 = 1;
    if (v1 != -1) ans -= v1;
    if (v2 != -1) ans -= v2;
    if (g1 != -1) ans -= g1;
    if (g2 != -1) ans -= g2;
    if (vg1 != -1) ans += vg1;
    if (vg2 != -1) ans += vg2;
    if (vg3 != -1) ans += vg3;
    if (vg4 != -1) ans += vg4;
    if (ans >= c) break;
  }
  cout << i;
  return 0;
}

```