#include <bits/stdc++.h>
using namespace std;
template <class Ta, class Tb>
inline Tb cast(Ta a) {
  stringstream ss;
  ss << a;
  Tb b;
  ss >> b;
  return b;
};
const double EPS = 1e-9;
const int INF = 1000000000;
const int MOD = 1000000007;
const int diri[8] = {-1, 0, 1, 0, -1, 1, 1, -1};
const int dirj[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int main() {
  int x;
  cin >> x;
  int n = 1;
  while (true) {
    if (n % 2 == 0) {
      int m = n / 2 - 1;
      int t = 4 * (((m + 1) / 2) * ((m + 1) / 2) + (m / 2) * (m / 2));
      if (x % 4 == 0 and x <= t) {
        cout << n << endl;
        return 0;
      }
    } else {
      int m = n / 2;
      for (int k = 0; k <= 1; ++k) {
        for (int i = 0; i <= (k ? m / 2 : (m + 1) / 2); ++i) {
          for (int j = 0; j <= (k ? m / 2 : (m + 1) / 2); ++j) {
            int q = k + 2 * i + 2 * j;
            if (q <= x) {
              q = x - q;
              int t = 4 * (((m + 1) / 2) * ((m + 1) / 2) + (m / 2) * (m / 2));
              if (k == 0 and (i == (m + 1) / 2 or j == (m + 1) / 2)) {
                t = 4 * (2 * (((m + 1) / 2) * (m / 2)));
              }
              if (q % 4 == 0 and q <= t) {
                cout << n << endl;
                return 0;
              }
            }
          }
        }
      }
    }
    ++n;
  }
}

### 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;
template <class Ta, class Tb>
inline Tb cast(Ta a) {
  stringstream ss;
  ss << a;
  Tb b;
  ss >> b;
  return b;
};
const double EPS = 1e-9;
const int INF = 1000000000;
const int MOD = 1000000007;
const int diri[8] = {-1, 0, 1, 0, -1, 1, 1, -1};
const int dirj[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int main() {
  int x;
  cin >> x;
  int n = 1;
  while (true) {
    if (n % 2 == 0) {
      int m = n / 2 - 1;
      int t = 4 * (((m + 1) / 2) * ((m + 1) / 2) + (m / 2) * (m / 2));
      if (x % 4 == 0 and x <= t) {
        cout << n << endl;
        return 0;
      }
    } else {
      int m = n / 2;
      for (int k = 0; k <= 1; ++k) {
        for (int i = 0; i <= (k ? m / 2 : (m + 1) / 2); ++i) {
          for (int j = 0; j <= (k ? m / 2 : (m + 1) / 2); ++j) {
            int q = k + 2 * i + 2 * j;
            if (q <= x) {
              q = x - q;
              int t = 4 * (((m + 1) / 2) * ((m + 1) / 2) + (m / 2) * (m / 2));
              if (k == 0 and (i == (m + 1) / 2 or j == (m + 1) / 2)) {
                t = 4 * (2 * (((m + 1) / 2) * (m / 2)));
              }
              if (q % 4 == 0 and q <= t) {
                cout << n << endl;
                return 0;
              }
            }
          }
        }
      }
    }
    ++n;
  }
}

```