#include <bits/stdc++.h>
using namespace std;
bool board[222][222];
int main(void) {
  int x = 0;
  scanf("%d", &x);
  for (int i = 1;; i++) {
    memset(board, 0, sizeof(board));
    int cx = x;
    if (cx & 1) {
      board[0][0] = true;
      cx--;
    }
    int canput = (i - 1 + (x % 2 == 0)) / 2;
    if (cx % 4 == 2) {
      if (canput < 1) continue;
      if (!board[0][0])
        board[0][1] = true;
      else
        board[0][2] = true;
    }
    int maxX = 0;
    if (board[0][0])
      maxX = ((i - 1) * (i - 1) + 1) / 2;
    else
      maxX = (i - 1) * (i - 1) / 2;
    if (maxX * 4 + canput * 4 >= cx) {
      printf("%d\n", 2 * i - 1);
      break;
    }
  }
  while (getchar() != EOF)
    ;
  return 0;
}

### Prompt

Generate a cpp solution to 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;
bool board[222][222];
int main(void) {
  int x = 0;
  scanf("%d", &x);
  for (int i = 1;; i++) {
    memset(board, 0, sizeof(board));
    int cx = x;
    if (cx & 1) {
      board[0][0] = true;
      cx--;
    }
    int canput = (i - 1 + (x % 2 == 0)) / 2;
    if (cx % 4 == 2) {
      if (canput < 1) continue;
      if (!board[0][0])
        board[0][1] = true;
      else
        board[0][2] = true;
    }
    int maxX = 0;
    if (board[0][0])
      maxX = ((i - 1) * (i - 1) + 1) / 2;
    else
      maxX = (i - 1) * (i - 1) / 2;
    if (maxX * 4 + canput * 4 >= cx) {
      printf("%d\n", 2 * i - 1);
      break;
    }
  }
  while (getchar() != EOF)
    ;
  return 0;
}

```