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;
long long arr[100][3];
void fill() {
  for (int i = 1; i < 100; i += 2) {
    long long x = (i - 1) / 2.0;
    long long y = ((x * x) + 1) / 2.0;
    arr[i][0] = y;
    x = i / 2.0;
    if (x % 2 == 1) x--;
    arr[i][1] = x;
    arr[i][2] = 1;
  }
}
bool ch(int n, int x) {
  if (n % 2 == 1) n--;
  if (n == 0) return 1;
  int y = n / 4;
  n -= min((long long)arr[x][0] * 4, (long long)y * 4);
  if (n == 0) return 1;
  y = n / 2;
  n -= min((long long)arr[x][1] * 2, (long long)y * 2);
  return n == 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  if (n == 2) {
    cout << 3;
    return 0;
  }
  fill();
  int res = -1;
  for (int i = 99; i > 0; i -= 2) {
    if (ch(n, i)) {
      res = i;
    }
  }
  cout << res << endl;
  return 0;
}
