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;
void prime(long long int a, long long int &ans, vector<long long int> &v) {
  for (long long int i = 2; i * i <= a; i++) {
    while (a % i == 0) {
      ans++;
      if (v.size() < 2) v.push_back(i);
      a = a / i;
    }
  }
  if (a > 1) ans++;
  if (a > 1 && v.size() < 2) v.push_back(a);
}
long long int mod;
long long int expmod(long long int a, long long int b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long int m1 = expmod(a, b / 2) % mod;
  if (b % 2) return ((m1 * m1) % mod * a % mod) % mod;
  return (m1 * m1) % mod;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  int x;
  cin >> x;
  int ans = 0;
  for (int i = 1; i < 101; i += 2) {
    int u = i / 2;
    int v = i / 2 + 1;
    int e = u * u + v * v;
    if (x <= e) {
      ans = i;
      break;
    }
  }
  if (x == 3)
    cout << 5;
  else
    cout << ans;
}
