#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long power(long long a, long long b, long long m = 1000000007) {
  long long answer = 1;
  while (b) {
    if (b & 1) answer = (answer * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return answer;
}
using namespace std;
void ctrl() { cout << "Control" << endl; }
long long make_num(string p) {
  stringstream geek(p);
  long long x = 0;
  geek >> x;
  return x;
}
string make_str(long long x) {
  ostringstream str1;
  str1 << x;
  string geek = str1.str();
  return geek;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  long long n;
  cin >> n;
  if (n < 4) {
    if (n == 1)
      cout << 1;
    else if (n == 2)
      cout << 3;
    else if (n == 3)
      cout << 5;
    return 0;
  }
  for (long long i = 1; i < 500; i += 2) {
    long long ans = 0;
    long long h = (i - 1) / 2;
    for (long long j = 1; j <= h; j++) {
      if (h % 2 == 0) {
        if (j % 2)
          ans += (h + 1) / 2;
        else
          ans += h / 2;
      } else {
        if (j % 2)
          ans += (h + 1) / 2;
        else
          ans += h / 2;
      }
    }
    ans = ans * 4;
    if (h % 2 == 0) {
      ans += 4 * (h / 2) + 1;
    } else {
      ans += 4 * (h / 2) + 1;
    }
    if (ans >= n) {
      cout << i << '\n';
      return 0;
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long power(long long a, long long b, long long m = 1000000007) {
  long long answer = 1;
  while (b) {
    if (b & 1) answer = (answer * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return answer;
}
using namespace std;
void ctrl() { cout << "Control" << endl; }
long long make_num(string p) {
  stringstream geek(p);
  long long x = 0;
  geek >> x;
  return x;
}
string make_str(long long x) {
  ostringstream str1;
  str1 << x;
  string geek = str1.str();
  return geek;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  long long n;
  cin >> n;
  if (n < 4) {
    if (n == 1)
      cout << 1;
    else if (n == 2)
      cout << 3;
    else if (n == 3)
      cout << 5;
    return 0;
  }
  for (long long i = 1; i < 500; i += 2) {
    long long ans = 0;
    long long h = (i - 1) / 2;
    for (long long j = 1; j <= h; j++) {
      if (h % 2 == 0) {
        if (j % 2)
          ans += (h + 1) / 2;
        else
          ans += h / 2;
      } else {
        if (j % 2)
          ans += (h + 1) / 2;
        else
          ans += h / 2;
      }
    }
    ans = ans * 4;
    if (h % 2 == 0) {
      ans += 4 * (h / 2) + 1;
    } else {
      ans += 4 * (h / 2) + 1;
    }
    if (ans >= n) {
      cout << i << '\n';
      return 0;
    }
  }
  return 0;
}

```