#include <bits/stdc++.h>
using namespace std;
void adskiy_razgon() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mxN = 1e+9 + 7;
const long long INF = 1e+18 + 7;
long long nod(long long a, long long b) {
  if (b > a) {
    swap(a, b);
  }
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nok(long long a, long long b) { return a * b / nod(a, b); }
void sp(long long a, double b) { cout << fixed << setprecision(a) << b; }
long long binpow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 1) {
    return binpow(a, n - 1) * a;
  } else {
    long long b = binpow(a, n / 2);
    return b * b;
  }
}
void solve() {
  long long n;
  cin >> n;
  if (n == 1) {
    cout << 1;
    return;
  }
  if (n == 3) {
    cout << 5;
    return;
  }
  if (n == 2) {
    cout << 3;
    return;
  }
  if (n == 4) {
    cout << 3;
    return;
  }
  for (int i = 1; i <= 10020; i += 2) {
    if ((i * i + 1) / 2 >= n) {
      cout << i;
      return;
    }
  }
}
int main() {
  adskiy_razgon();
  long long t = 1;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
using namespace std;
void adskiy_razgon() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mxN = 1e+9 + 7;
const long long INF = 1e+18 + 7;
long long nod(long long a, long long b) {
  if (b > a) {
    swap(a, b);
  }
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nok(long long a, long long b) { return a * b / nod(a, b); }
void sp(long long a, double b) { cout << fixed << setprecision(a) << b; }
long long binpow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 1) {
    return binpow(a, n - 1) * a;
  } else {
    long long b = binpow(a, n / 2);
    return b * b;
  }
}
void solve() {
  long long n;
  cin >> n;
  if (n == 1) {
    cout << 1;
    return;
  }
  if (n == 3) {
    cout << 5;
    return;
  }
  if (n == 2) {
    cout << 3;
    return;
  }
  if (n == 4) {
    cout << 3;
    return;
  }
  for (int i = 1; i <= 10020; i += 2) {
    if ((i * i + 1) / 2 >= n) {
      cout << i;
      return;
    }
  }
}
int main() {
  adskiy_razgon();
  long long t = 1;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

```