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;
vector<int> sieve(int n) {
  vector<bool> prime(n + 1, true);
  prime[0] = false;
  prime[1] = false;
  int m = sqrt(n);
  vector<int> ans;
  for (int p = 2; p <= m; p++) {
    if (prime[p]) {
      for (int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
  for (int i = 0; i < n; i++)
    if (prime[i]) ans.push_back(i);
  return ans;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int mod = 1e9 + 7;
int power(int a, long long b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1LL * res * a;
    a = 1LL * a * a;
    b >>= 1;
  }
  return res;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long min(int a, long long b) {
  if (a > b) return b;
  return a;
}
long long max(int a, long long b) {
  if (a > b) return a;
  return b;
}
long long myXOR(long long x, long long y) { return (x | y) & (~x | ~y); }
int conti(int x) {
  if (x < 1) return 0;
  if (x % 4 == 0) return x;
  if (x % 4 == 1) return 1;
  if (x % 4 == 2) return x + 1;
  if (x % 4 == 3) return 0;
}
bool ispfsq(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
vector<int> adj[200005];
bool visited[200005];
int ch[100005], dep[100005];
long long t = 0;
double first = INT_MIN;
vector<int> ans;
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  if (n == 3) {
    cout << 5;
    return;
  }
  for (int i = 1; i < 100; i += 2) {
    if (i * i / 2 + 1 >= n) {
      cout << i;
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int mul = 0;
  if (mul == 1) {
    int t;
    cin >> t;
    while (t--) {
      solve();
      cout << "\n";
    }
  } else {
    solve();
  }
}
