#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
const int len = 2097155;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    int x;
    cin >> x;
    if (x == 3) {
      cout << 5;
      return 0;
    }
    int p = 1;
    while (power(p, 2) + 1 < 2 * x) p += 2;
    cout << p;
    return 0;
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int mod = 998244353;
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
const int len = 2097155;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    int x;
    cin >> x;
    if (x == 3) {
      cout << 5;
      return 0;
    }
    int p = 1;
    while (power(p, 2) + 1 < 2 * x) p += 2;
    cout << p;
    return 0;
  }
}

```