#include <bits/stdc++.h>
using namespace std;
int count(int n) {
  if (n <= 0) return 0;
  if (n % 2 == 1) {
    return count(n - 1) + 1 + 2 * ((n - 1) / 2);
  } else
    return n * n / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  int x, best;
  cin >> x;
  if (x == 3) {
    cout << 5 << endl;
    return 0;
  }
  for (int n = 1; n <= 2000; n += 2) {
    bool ok = false;
    if (n % 2 == 0) {
      if (x % 4 != 0) continue;
      int m = n / 2 - 1;
      if (m <= 0) continue;
      if (m % 2 == 0) {
        best = 4 * count(m);
        if (best >= x) ok = true;
      } else {
        best = 4 * count(m);
        if (best >= x) ok = true;
      }
    } else {
      int m = (n - 1) / 2;
      int y = x;
      int A = count(m), B = (m % 2 == 0 ? m / 2 : (m - 1) / 2), C = B;
      if (y % 2 == 1) y--;
      if (4 * A >= y) ok = true;
      y -= 4 * A;
      if (2 * B >= y) ok = true;
      y -= 2 * B;
      if (2 * C >= y) ok = true;
    }
    if (ok) {
      cout << n << endl;
      return 0;
    }
  }
  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;
int count(int n) {
  if (n <= 0) return 0;
  if (n % 2 == 1) {
    return count(n - 1) + 1 + 2 * ((n - 1) / 2);
  } else
    return n * n / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  int x, best;
  cin >> x;
  if (x == 3) {
    cout << 5 << endl;
    return 0;
  }
  for (int n = 1; n <= 2000; n += 2) {
    bool ok = false;
    if (n % 2 == 0) {
      if (x % 4 != 0) continue;
      int m = n / 2 - 1;
      if (m <= 0) continue;
      if (m % 2 == 0) {
        best = 4 * count(m);
        if (best >= x) ok = true;
      } else {
        best = 4 * count(m);
        if (best >= x) ok = true;
      }
    } else {
      int m = (n - 1) / 2;
      int y = x;
      int A = count(m), B = (m % 2 == 0 ? m / 2 : (m - 1) / 2), C = B;
      if (y % 2 == 1) y--;
      if (4 * A >= y) ok = true;
      y -= 4 * A;
      if (2 * B >= y) ok = true;
      y -= 2 * B;
      if (2 * C >= y) ok = true;
    }
    if (ok) {
      cout << n << endl;
      return 0;
    }
  }
  return 0;
}

```