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;
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
int n, t, m, a, b, c, ans;
typedef struct taolao {
  int l, r;
};
taolao T[20];
int main() {
  scanf("%d", &a);
  if (a == 1)
    ans = 1;
  else if (a == 3)
    ans = 5;
  else {
    int tmp = 1;
    int tmp2 = 1;
    while (tmp2 < a) {
      tmp += 2;
      tmp2 = (tmp * tmp + 1) / 2;
    }
    ans = tmp;
  }
  printf("%d\n", ans);
  return 0;
}
