#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  long long last = 0;
  for (int i = 1;; i++) {
    if (i & 1) {
      long long cnt = (1ll * i * i) / 2 + 1;
      if (cnt >= n) {
        int r = n % 4;
        if (r == 2 and i >= 3) {
          cout << i;
          return;
        } else if (r == 3 and i >= 5) {
          cout << i;
          return;
        } else if (r == 1 or r == 0) {
          cout << i;
          return;
        }
      }
    } else {
      if (i != 2 and (n % 4 == 0)) {
        long long c = (i / 4 - 1);
        long long cnt = last + 4 + c * 8;
        last = cnt;
        if (cnt >= n) {
          cout << i;
          return;
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
}

### 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 MOD = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  long long last = 0;
  for (int i = 1;; i++) {
    if (i & 1) {
      long long cnt = (1ll * i * i) / 2 + 1;
      if (cnt >= n) {
        int r = n % 4;
        if (r == 2 and i >= 3) {
          cout << i;
          return;
        } else if (r == 3 and i >= 5) {
          cout << i;
          return;
        } else if (r == 1 or r == 0) {
          cout << i;
          return;
        }
      }
    } else {
      if (i != 2 and (n % 4 == 0)) {
        long long c = (i / 4 - 1);
        long long cnt = last + 4 + c * 8;
        last = cnt;
        if (cnt >= n) {
          cout << i;
          return;
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
}

```