#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 10;
const long long mod = 1e9 + 7;
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  long long k = pw(a, b / 2);
  k = (k * k) % mod;
  if (b & 1) k = (k * a) % mod;
  return k;
}
long long n, m;
long long a[maxn];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  if (x == 1) return cout << 1, 0;
  if (x == 2) return cout << 3, 0;
  if (x == 3) return cout << 5, 0;
  for (long long n = 1; n <= 1000; n++) {
    if (n & 1)
      if (x <= (n * n + 1) / 2) return cout << n, 0;
    if (x % 4 == 0)
      if (n % 2 == 0)
        if (x <= (n - 2) * (n - 2) / 2) return cout << n, 0;
  }
}

### Prompt

Your task is to create a CPP solution 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>
using namespace std;
const long long maxn = 2e5 + 10;
const long long mod = 1e9 + 7;
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  long long k = pw(a, b / 2);
  k = (k * k) % mod;
  if (b & 1) k = (k * a) % mod;
  return k;
}
long long n, m;
long long a[maxn];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  if (x == 1) return cout << 1, 0;
  if (x == 2) return cout << 3, 0;
  if (x == 3) return cout << 5, 0;
  for (long long n = 1; n <= 1000; n++) {
    if (n & 1)
      if (x <= (n * n + 1) / 2) return cout << n, 0;
    if (x % 4 == 0)
      if (n % 2 == 0)
        if (x <= (n - 2) * (n - 2) / 2) return cout << n, 0;
  }
}

```