#include <bits/stdc++.h>
using namespace std;
int i, j, n;
bool ok(int n, int x) {
  int i, j, k;
  int slice = n >> 1;
  int init = (slice + 1) * (slice + 1) + slice * slice;
  int init2 = 2 * (slice + 1) * slice;
  for (i = 0; i <= slice * slice / 2; ++i)
    for (j = 0; j <= 2 * (slice - 1); ++j)
      for (k = 0; k <= 1; ++k)
        if (init - i * 4 - 2 * j - 1 * k == x) return true;
  for (i = 0; i <= slice * slice / 2; ++i)
    for (j = 0; j <= 2 * (slice); ++j)
      if (init2 - i * 4 - 2 * j == x) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (i = 3;; i += 2) {
    if (ok(i, n)) break;
  }
  printf("%d\n", i);
  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;
int i, j, n;
bool ok(int n, int x) {
  int i, j, k;
  int slice = n >> 1;
  int init = (slice + 1) * (slice + 1) + slice * slice;
  int init2 = 2 * (slice + 1) * slice;
  for (i = 0; i <= slice * slice / 2; ++i)
    for (j = 0; j <= 2 * (slice - 1); ++j)
      for (k = 0; k <= 1; ++k)
        if (init - i * 4 - 2 * j - 1 * k == x) return true;
  for (i = 0; i <= slice * slice / 2; ++i)
    for (j = 0; j <= 2 * (slice); ++j)
      if (init2 - i * 4 - 2 * j == x) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (i = 3;; i += 2) {
    if (ok(i, n)) break;
  }
  printf("%d\n", i);
  return 0;
}

```