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 N, Sharp;
inline bool Black(int S) {
  int CS;
  if (N % 2 == 0) {
    int M = N / 2 - 1;
    CS = 4 * (M * M / 2 + (M & 1));
    S = max(S % 4, S - CS);
    if (!S) return true;
  } else {
    int M = N / 2;
    CS = 4 * (M * M / 2 + (M & 1));
    S = max(S % 4, S - CS);
    if (!S) return true;
    if ((M & 1) and S > N - 2) return false;
    if ((!(M & 1)) and S > N) return false;
    if (S % 2 == 0) return true;
    if ((S - 1) % 2 == 0) return true;
  }
  return false;
}
inline bool White(int S) {
  int CS;
  if (N % 2 == 0) return false;
  int M = N / 2;
  CS = 4 * (M * M / 2);
  S = max(S % 4, S - CS);
  if (!S) return true;
  if ((M & 1) and S > N + 1) return false;
  if ((!(M & 1)) and S > N - 1) return false;
  if (S % 2 == 0) return true;
  return false;
}
inline void Solve() {
  for (N = 1;; ++N) {
    if (Black(Sharp)) return;
    if (White(Sharp)) return;
  }
}
inline void Read() { scanf("%d", &Sharp); }
inline void Print() { printf("%d\n", N); }
int main() {
  Read();
  Solve();
  Print();
  return 0;
}
