#include <bits/stdc++.h>
using namespace std;
int solve(int n) {
  if (n == 1) return 1;
  if (n == 2) return 3;
  if (n == 3) return 5;
  if (n == 4) return 3;
  if (n % 4 == 0) {
    for (int i = (4); i < (110); ++i) {
      if (i % 2 == 0) {
        int r = (i - 2) / 2;
        int can = 4 * ((r * r + 1) / 2);
        if (can >= n) return i;
      } else {
        int r = (i - 1) / 2;
        int can = 4 * ((r * r + 1) / 2);
        int app = 4 * (r / 2);
        app++;
        can += app;
        if (n <= can) return i;
      }
    }
  } else {
    for (int i = 3; i < 111; i += 2) {
      int r = (i - 1) / 2;
      int can = 4 * ((r * r + 1) / 2);
      int app = 4 * (r / 2);
      app++;
      can += app;
      if (n <= can) return i;
    }
  }
  return 0;
}
int main() {
  int x;
  while (cin >> x) {
    cout << solve(x) << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int solve(int n) {
  if (n == 1) return 1;
  if (n == 2) return 3;
  if (n == 3) return 5;
  if (n == 4) return 3;
  if (n % 4 == 0) {
    for (int i = (4); i < (110); ++i) {
      if (i % 2 == 0) {
        int r = (i - 2) / 2;
        int can = 4 * ((r * r + 1) / 2);
        if (can >= n) return i;
      } else {
        int r = (i - 1) / 2;
        int can = 4 * ((r * r + 1) / 2);
        int app = 4 * (r / 2);
        app++;
        can += app;
        if (n <= can) return i;
      }
    }
  } else {
    for (int i = 3; i < 111; i += 2) {
      int r = (i - 1) / 2;
      int can = 4 * ((r * r + 1) / 2);
      int app = 4 * (r / 2);
      app++;
      can += app;
      if (n <= can) return i;
    }
  }
  return 0;
}
int main() {
  int x;
  while (cin >> x) {
    cout << solve(x) << endl;
  }
  return 0;
}

```