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;
const int inf = 1000000000;
const int MOD = 1000000007;
bool f(int n, int cnt) {
  int m = n / 2;
  if (m % 2 == 0) {
    return 2 * m * m + 2 * m + 1 >= cnt;
  } else {
    int k1 = (m * m) / 2, k2 = (m * m) - k1;
    int num1 = 4 * k2 + 4 * (m / 2) + 1;
    int num2 = 4 * k1 + 4 * (m - m / 2);
    if (cnt % 2 == 1) {
      return num1 >= cnt;
    } else {
      return max(num1 - 1, num2) >= cnt;
    }
  }
}
bool F(int n, int cnt) {
  int h = (n * n + 1) / 2;
  int hh = (h - 1) / 4;
  if (cnt + 4 * hh < h) return false;
  return h >= cnt;
}
int main() {
  int n;
  cin >> n;
  if (n == 3) {
    cout << "5" << endl;
    return 0;
  }
  for (int i = 1;; i += 2)
    if (F(i, n)) {
      cout << i << endl;
      return 0;
    }
  return 0;
}
