#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using VI = vector<int>;
using VL = vector<long long>;
using VVI = vector<VI>;
using VVL = vector<VL>;
using MII = map<int, int>;
using MIVI = map<int, VI>;
using MLL = map<LL, LL>;
int READ_INT() {
  int temp;
  cin >> temp;
  return temp;
}
LL READ_LONG() {
  LL temp;
  cin >> temp;
  return temp;
}
const int MOD = int(1e9) + 7;
const int INF = 1e9 + 5;
const double PI = acos(-1.0);
const double EPS = 1e-9;
int main() {
  ios::sync_with_stdio(false);
  int x = READ_INT();
  if (x == 3) {
    cout << 5 << "\n";
    return 0;
  }
  for (int n = 1; n <= 100; n += 2) {
    if ((n * n + 1) / 2 >= x) {
      cout << n << "\n";
      break;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
using LL = long long;
using VI = vector<int>;
using VL = vector<long long>;
using VVI = vector<VI>;
using VVL = vector<VL>;
using MII = map<int, int>;
using MIVI = map<int, VI>;
using MLL = map<LL, LL>;
int READ_INT() {
  int temp;
  cin >> temp;
  return temp;
}
LL READ_LONG() {
  LL temp;
  cin >> temp;
  return temp;
}
const int MOD = int(1e9) + 7;
const int INF = 1e9 + 5;
const double PI = acos(-1.0);
const double EPS = 1e-9;
int main() {
  ios::sync_with_stdio(false);
  int x = READ_INT();
  if (x == 3) {
    cout << 5 << "\n";
    return 0;
  }
  for (int n = 1; n <= 100; n += 2) {
    if ((n * n + 1) / 2 >= x) {
      cout << n << "\n";
      break;
    }
  }
  return 0;
}

```