#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-12;
inline int read() {
  static char ch;
  bool sgn = false;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') sgn = true;
  int res = ch - 48;
  while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
  return sgn ? -res : res;
}
const int N = 1e5 + 5;
int ans[N];
int main() {
  int x = read();
  fill(ans, ans + 200, 1 << 30);
  for (int i = 2; i <= 100; i += 2) {
    int j = 2 * i - 4;
    while (j > 0) {
      ans[j] = (ans[j] < i ? ans[j] : i);
      j -= 4;
    }
  }
  for (int i = 1; i <= 100; i++) {
    if (i % 2) {
      for (int j = 1; j <= (i * i + 1) / 2; j++)
        ans[j] = (ans[j] < i ? ans[j] : i);
    }
  }
  ans[3] = 5;
  ans[1] = 1;
  ans[2] = 3;
  ans[4] = 3;
  ans[5] = 3;
  cout << ans[x];
}

### Prompt

Please provide a Cpp coded solution 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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-12;
inline int read() {
  static char ch;
  bool sgn = false;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') sgn = true;
  int res = ch - 48;
  while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
  return sgn ? -res : res;
}
const int N = 1e5 + 5;
int ans[N];
int main() {
  int x = read();
  fill(ans, ans + 200, 1 << 30);
  for (int i = 2; i <= 100; i += 2) {
    int j = 2 * i - 4;
    while (j > 0) {
      ans[j] = (ans[j] < i ? ans[j] : i);
      j -= 4;
    }
  }
  for (int i = 1; i <= 100; i++) {
    if (i % 2) {
      for (int j = 1; j <= (i * i + 1) / 2; j++)
        ans[j] = (ans[j] < i ? ans[j] : i);
    }
  }
  ans[3] = 5;
  ans[1] = 1;
  ans[2] = 3;
  ans[4] = 3;
  ans[5] = 3;
  cout << ans[x];
}

```