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;
vector<int> odd_list;
void build() {
  for (int i = 1; i < 100; i += 2) {
    odd_list.push_back(i);
  }
}
int check(int odd, int x) {
  int a, b, rem, aux = (odd - 1) / 2;
  for (int i = 1; i <= (odd / 2) + 1; i++) {
    a = i;
    b = i - 1;
    rem = x - (aux * a + aux * b);
    if (rem <= i) return true;
  }
  return false;
}
int BS(int x) {
  int lo = 0, hi = odd_list.size() - 1, ans;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    int odd = odd_list[mid];
    if (check(odd, x)) {
      ans = odd;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int x;
  cin >> x;
  build();
  int ans = BS(x);
  cout << (x == 3 ? 5 : ans) << '\n';
  return 0;
}
