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;
map<long long, long long> m1;
bool f(long long n, long long x) {
  long long i, j;
  if (n % 2 == 0) return false;
  long long mat[n][n];
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      mat[i][j] = 0;
    }
  }
  long long c1 = 0LL;
  for (i = 0; i < n; i += 2) {
    for (j = 0; j < n; j += 2) {
      mat[j][i] = 1;
      c1++;
    }
  }
  if (c1 == x) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long x;
  cin >> x;
  long long low = 1, high = 200;
  long long ans = INT_MAX;
  long long i;
  if (x == 3) {
    cout << 5;
    return 0;
  }
  for (int i = 1;; i += 2) {
    int mi = (i / 2) * (i / 2) + ((i + 1) / 2) * ((i + 1) / 2);
    if (x <= mi) {
      cout << i << '\n';
      return 0;
    }
  }
  return 0;
}
