#include <bits/stdc++.h>
using namespace std;
int x;
bool F(int n) {
  int m = n / 2;
  vector<int> T;
  int c2 = 0, c4 = 0, c = 0;
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (n); ++(j)) {
      int a = ((i % 2) == (j % 2));
      if (a == 0) continue;
      c++;
      if (i == m && j == m)
        T.push_back(1);
      else if (i == m || j == m)
        c2++;
      else
        c4++;
    }
  if (c < x) return false;
  for (int(i) = (0); (i) < (c2 / 2); ++(i)) T.push_back(2);
  for (int(j) = (0); (j) < (c4 / 4); ++(j)) T.push_back(4);
  reverse((T).begin(), (T).end());
  for (int(i) = (0); (i) < (T.size()); ++(i))
    if (c - T[i] >= x) c -= T[i];
  return (c == x);
}
bool FF(int n) {
  int m = n / 2;
  vector<int> T;
  int c2 = 0, c4 = 0, c = 0;
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (n); ++(j)) {
      int a = ((i % 2) != (j % 2));
      if (a == 0) continue;
      c++;
      if (i == m && j == m)
        T.push_back(1);
      else if (i == m || j == m)
        c2++;
      else
        c4++;
    }
  if (c < x) return false;
  for (int(i) = (0); (i) < (c2 / 2); ++(i)) T.push_back(2);
  for (int(j) = (0); (j) < (c4 / 4); ++(j)) T.push_back(4);
  reverse((T).begin(), (T).end());
  for (int(i) = (0); (i) < (T.size()); ++(i))
    if (c - T[i] >= x) c -= T[i];
  return (c == x);
}
int main() {
  cin >> x;
  for (int i = 1; i <= 1000000000; i += 2) {
    if (F(i) || FF(i)) {
      cout << i << endl;
      return 0;
    }
  }
  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;
int x;
bool F(int n) {
  int m = n / 2;
  vector<int> T;
  int c2 = 0, c4 = 0, c = 0;
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (n); ++(j)) {
      int a = ((i % 2) == (j % 2));
      if (a == 0) continue;
      c++;
      if (i == m && j == m)
        T.push_back(1);
      else if (i == m || j == m)
        c2++;
      else
        c4++;
    }
  if (c < x) return false;
  for (int(i) = (0); (i) < (c2 / 2); ++(i)) T.push_back(2);
  for (int(j) = (0); (j) < (c4 / 4); ++(j)) T.push_back(4);
  reverse((T).begin(), (T).end());
  for (int(i) = (0); (i) < (T.size()); ++(i))
    if (c - T[i] >= x) c -= T[i];
  return (c == x);
}
bool FF(int n) {
  int m = n / 2;
  vector<int> T;
  int c2 = 0, c4 = 0, c = 0;
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (n); ++(j)) {
      int a = ((i % 2) != (j % 2));
      if (a == 0) continue;
      c++;
      if (i == m && j == m)
        T.push_back(1);
      else if (i == m || j == m)
        c2++;
      else
        c4++;
    }
  if (c < x) return false;
  for (int(i) = (0); (i) < (c2 / 2); ++(i)) T.push_back(2);
  for (int(j) = (0); (j) < (c4 / 4); ++(j)) T.push_back(4);
  reverse((T).begin(), (T).end());
  for (int(i) = (0); (i) < (T.size()); ++(i))
    if (c - T[i] >= x) c -= T[i];
  return (c == x);
}
int main() {
  cin >> x;
  for (int i = 1; i <= 1000000000; i += 2) {
    if (F(i) || FF(i)) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}

```