#include <bits/stdc++.h>
using namespace std;
const int N = 110005;
const int mod = 1e9 + 7;
int ans;
int main() {
  int x;
  cin >> x;
  int z = x;
  while (x) {
    int y;
    if (x % 2 == 0) {
      y = x * (x / 2);
    } else {
      y = (x / 2) * (x / 2) + ((x + 1) * (x + 1)) / 4;
    }
    if (y >= z) ans = x;
    x--;
  }
  if (z == 3 || z == 2) ans = 5;
  if (z == 2) ans = 3;
  if (ans % 2 == 0) ans++;
  cout << ans;
}

### Prompt

Create a solution in CPP for 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;
const int N = 110005;
const int mod = 1e9 + 7;
int ans;
int main() {
  int x;
  cin >> x;
  int z = x;
  while (x) {
    int y;
    if (x % 2 == 0) {
      y = x * (x / 2);
    } else {
      y = (x / 2) * (x / 2) + ((x + 1) * (x + 1)) / 4;
    }
    if (y >= z) ans = x;
    x--;
  }
  if (z == 3 || z == 2) ans = 5;
  if (z == 2) ans = 3;
  if (ans % 2 == 0) ans++;
  cout << ans;
}

```