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>
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
long double PI = 3.14159265358979323846;
const int modo = 1e9 + 7;
const long long inf = 1e13;
const long long ms = (1e6) + 5;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n;
  cin >> n;
  int i = 1;
  int a;
  int b;
  int aa;
  int bb;
  int tt;
  if (n == 3) {
    cout << "5";
    return 0;
  }
  while (true) {
    a = i / 2;
    b = (i + 1) / 2;
    tt = (a * a) + (b * b);
    if (n <= tt) {
      cout << i;
      break;
    }
    i += 2;
  }
  return 0;
}
