#include <bits/stdc++.h>
using namespace std;
int i, x;
bool can(int N, int x) {
  if (N == 1) return (x <= 1);
  if (N == 2) return (x <= 0);
  if (N == 3) return (x <= 5 && x != 3);
  if (N % 2 == 1) return (x <= (N * N + 1) / 2);
  if (x % 4 != 0) return false;
  int n = N / 2 - 1;
  int ans = (n * n + 1) / 2;
  ans *= 4;
  return (x <= ans);
}
int main() {
  scanf("%d", &x);
  for (i = 1;; i++)
    if (can(i, x)) break;
  printf("%d\n", i);
  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;
int i, x;
bool can(int N, int x) {
  if (N == 1) return (x <= 1);
  if (N == 2) return (x <= 0);
  if (N == 3) return (x <= 5 && x != 3);
  if (N % 2 == 1) return (x <= (N * N + 1) / 2);
  if (x % 4 != 0) return false;
  int n = N / 2 - 1;
  int ans = (n * n + 1) / 2;
  ans *= 4;
  return (x <= ans);
}
int main() {
  scanf("%d", &x);
  for (i = 1;; i++)
    if (can(i, x)) break;
  printf("%d\n", i);
  return 0;
}

```