#include <bits/stdc++.h>
using namespace std;
int solve(int x) {
  if (x % 2) {
    const int y = x - 1;
    for (int n = 1;; n += 2) {
      const int h = n / 2;
      if (y % 4 == 2 && h <= 1) continue;
      const int m = h * (h + 1) / 2;
      if (m * 4 >= y) return n;
    }
  } else {
    for (int n = 1;; n++) {
      if (n % 2) {
        const int h = n / 2;
        const int m = h * (h + 1) / 2;
        if (m * 4 >= x) return n;
      } else {
        const int h = n / 2 - 1;
        const int m = h * h / 2 + h % 2;
        if (x % 4 != 0) continue;
        if (m * 4 >= x) return n;
      }
    }
  }
  return -1;
}
int main() {
  int x;
  scanf("%d", &x);
  printf("%d\n", solve(x));
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
int solve(int x) {
  if (x % 2) {
    const int y = x - 1;
    for (int n = 1;; n += 2) {
      const int h = n / 2;
      if (y % 4 == 2 && h <= 1) continue;
      const int m = h * (h + 1) / 2;
      if (m * 4 >= y) return n;
    }
  } else {
    for (int n = 1;; n++) {
      if (n % 2) {
        const int h = n / 2;
        const int m = h * (h + 1) / 2;
        if (m * 4 >= x) return n;
      } else {
        const int h = n / 2 - 1;
        const int m = h * h / 2 + h % 2;
        if (x % 4 != 0) continue;
        if (m * 4 >= x) return n;
      }
    }
  }
  return -1;
}
int main() {
  int x;
  scanf("%d", &x);
  printf("%d\n", solve(x));
  return 0;
}

```