#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const long long mod = 1e9 + 7;
const int oo = 1e9;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int x;
int main() {
  scanf("%d", &x);
  if (x == 3) {
    printf("5");
    return 0;
  }
  for (int i = 1; i < 1000; i += 2) {
    int x1 = (i / 2) * (i / 2), x2 = (i / 2) * ((i + 1) / 2), x3 = 0;
    if (i % 2 != 0) x3 += (i / 2) + 1;
    if (x1 + x2 + x3 >= x) {
      printf("%d", i);
      return 0;
    }
  }
}

### Prompt

Create a solution in cpp for 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 int N = 1e6 + 5;
const long long mod = 1e9 + 7;
const int oo = 1e9;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int x;
int main() {
  scanf("%d", &x);
  if (x == 3) {
    printf("5");
    return 0;
  }
  for (int i = 1; i < 1000; i += 2) {
    int x1 = (i / 2) * (i / 2), x2 = (i / 2) * ((i + 1) / 2), x3 = 0;
    if (i % 2 != 0) x3 += (i / 2) + 1;
    if (x1 + x2 + x3 >= x) {
      printf("%d", i);
      return 0;
    }
  }
}

```