#include <bits/stdc++.h>
using namespace std;
const long long dx[] = {0, -1, 0, 1};
const long long dy[] = {-1, 0, 1, 0};
const long long INF = 2e17;
const long long N = 2e5 + 5;
const long long M = 1e9 + 7;
void mini(long long &a, long long b) { a = min(a, b); }
void maxi(long long &a, long long b) { a = max(a, b); }
void solve() {
  long long x;
  cin >> x;
  long long s = 1, e = 30, ans = 100;
  if (x == 1)
    ans = 1;
  else if (x == 3)
    ans = 5;
  else if (x == 2 || x == 4 || x == 5)
    ans = 3;
  else {
    for (long long i = 5; i < 30; i += 2) {
      if (x <= (i * i + 1) / 2) {
        ans = i;
        break;
      }
    }
  }
  cout << ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  ;
  long long tt = 1;
  while (tt--) {
    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;
const long long dx[] = {0, -1, 0, 1};
const long long dy[] = {-1, 0, 1, 0};
const long long INF = 2e17;
const long long N = 2e5 + 5;
const long long M = 1e9 + 7;
void mini(long long &a, long long b) { a = min(a, b); }
void maxi(long long &a, long long b) { a = max(a, b); }
void solve() {
  long long x;
  cin >> x;
  long long s = 1, e = 30, ans = 100;
  if (x == 1)
    ans = 1;
  else if (x == 3)
    ans = 5;
  else if (x == 2 || x == 4 || x == 5)
    ans = 3;
  else {
    for (long long i = 5; i < 30; i += 2) {
      if (x <= (i * i + 1) / 2) {
        ans = i;
        break;
      }
    }
  }
  cout << ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  ;
  long long tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

```