#include <bits/stdc++.h>
int sum(int x) {
  if (x % 2 == 0)
    return x * x / 2;
  else {
    int re = 0;
    for (int i = 1; i <= x; ++i) {
      if (i & 1)
        re += (x + 1) / 2;
      else
        re += x / 2;
    }
    return re;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int ans = n * 2 - 1;
  if (n & 1) --n;
  for (int L = 1; L <= 1000; ++L) {
    for (int x = 0; x <= L / 2 + L / 2; ++x) {
      int res = n - x * 2;
      if (res >= 0 && res % 4 == 0 && sum(L) >= res / 4) {
        ans = std::min(ans, L * 2 + 1);
      }
    }
  }
  printf("%d\n", ans);
  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>
int sum(int x) {
  if (x % 2 == 0)
    return x * x / 2;
  else {
    int re = 0;
    for (int i = 1; i <= x; ++i) {
      if (i & 1)
        re += (x + 1) / 2;
      else
        re += x / 2;
    }
    return re;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int ans = n * 2 - 1;
  if (n & 1) --n;
  for (int L = 1; L <= 1000; ++L) {
    for (int x = 0; x <= L / 2 + L / 2; ++x) {
      int res = n - x * 2;
      if (res >= 0 && res % 4 == 0 && sum(L) >= res / 4) {
        ans = std::min(ans, L * 2 + 1);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```