#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
struct node {
  long long int t, i, j, k;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, m, k, q, x, a, b, c, d, w;
  cin >> n;
  vector<int> ans(20), vec(200);
  ans[1] = 1;
  ans[0] = 0;
  vec[1] = 1;
  for (int i = 3; i < 20; i += 2) {
    ans[i] += ans[i - 2] + 2 * i - 2;
    if (ans[i] < 200) {
      vec[ans[i]] = i;
    }
  }
  for (int i = 198; i >= 0; i--) {
    if (!vec[i]) vec[i] = vec[i + 1];
  }
  vec[3] = 5;
  cout << vec[n] << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 mod = 1000000007;
struct node {
  long long int t, i, j, k;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, m, k, q, x, a, b, c, d, w;
  cin >> n;
  vector<int> ans(20), vec(200);
  ans[1] = 1;
  ans[0] = 0;
  vec[1] = 1;
  for (int i = 3; i < 20; i += 2) {
    ans[i] += ans[i - 2] + 2 * i - 2;
    if (ans[i] < 200) {
      vec[ans[i]] = i;
    }
  }
  for (int i = 198; i >= 0; i--) {
    if (!vec[i]) vec[i] = vec[i + 1];
  }
  vec[3] = 5;
  cout << vec[n] << "\n";
  return 0;
}

```