#include <bits/stdc++.h>
using namespace std;
bool valid(int n, int x) {
  if (n == 1) {
    return x == 1;
  }
  if (n == 2) {
    return false;
  }
  if (n == 3) {
    return x == 1 || x == 2 || x == 4 || x == 5;
  }
  if (n % 2 == 1) {
    int f = (1 + (n / 2) * (n / 2)) / 2;
    int t = (n / 4) * 2;
    x -= x % 2;
    if (x % 4 == 2) {
      t--;
      x -= 2;
    }
    f += t / 2;
    return f * 4 >= x;
  } else {
    int f = (1 + (n / 2 - 1) * (n / 2 - 1)) / 2;
    return f * 4 >= x && x % 4 == 0;
  }
}
int main() {
  int x;
  cin >> x;
  int n = 1;
  while (!valid(n, x)) {
    n++;
  }
  cout << n << '\n';
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.

Let's call matrix A clear if no two cells containing ones have a common side.

Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.

Let's define the sharpness of matrix A as the number of ones in it.

Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.

Input

The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.

Output

Print a single number — the sought value of n.

Examples

Input

4


Output

3


Input

9


Output

5

Note

The figure below shows the matrices that correspond to the samples:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool valid(int n, int x) {
  if (n == 1) {
    return x == 1;
  }
  if (n == 2) {
    return false;
  }
  if (n == 3) {
    return x == 1 || x == 2 || x == 4 || x == 5;
  }
  if (n % 2 == 1) {
    int f = (1 + (n / 2) * (n / 2)) / 2;
    int t = (n / 4) * 2;
    x -= x % 2;
    if (x % 4 == 2) {
      t--;
      x -= 2;
    }
    f += t / 2;
    return f * 4 >= x;
  } else {
    int f = (1 + (n / 2 - 1) * (n / 2 - 1)) / 2;
    return f * 4 >= x && x % 4 == 0;
  }
}
int main() {
  int x;
  cin >> x;
  int n = 1;
  while (!valid(n, x)) {
    n++;
  }
  cout << n << '\n';
  return 0;
}

```