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;
int main() {
  int n, x, res = 1;
  scanf("%d", &x);
  for (res = 1;; res++) {
    int s = 0;
    if (res % 2 == 0) {
      for (int i = 1; i * 2 < res; i++) {
        if (i % 2) {
          int p = res / 2;
          if (p % 2) p--;
          s += 2 * p;
        } else {
          int p = (res - 2) / 2;
          if (p % 2) p--;
          s += 2 * p;
        }
      }
      if (s >= x && (s - x) % 4 == 0) break;
    } else {
      int q = (res + 1) / 2, p;
      for (int i = 1; i * 2 <= res + 1; i++) {
        if (i % 2 == 0)
          p = q - 1;
        else
          p = q;
        if ((res + 1) / 2 == i)
          s += p;
        else
          s += 2 * p;
      }
      int ond1 = -1 + (res - 1) / 2 * 2, ond2 = (res + 1) / 2 * 2;
      if (s >= x && (x % 4) <= ond1) break;
      if (s > x && (x % 4) <= ond2 && x % 2 == 0) break;
      if (s >= x && (s - x) % 4 == 0) break;
      if (s - 1 >= x && (s - 1 - x) % 4 == 0) break;
    }
  }
  cout << res << endl;
  return 0;
}
