#include <bits/stdc++.h>
using namespace std;
int ans[128];
int main() {
  for (int i = 1; i <= 100; i++) {
    ans[i] = 10000;
    if (i % 4 == 0) {
      for (int n = 4;; n += 2) {
        int hn = n / 2 - 1;
        int maxval = 0;
        if (hn % 2) {
          maxval = 4 * ((hn * hn + 1) / 2);
        } else {
          maxval = 4 * (hn * hn / 2);
        }
        if (maxval >= i) {
          ans[i] = min(ans[i], n);
          break;
        }
      }
    }
    for (int n = 5;; n += 2) {
      int maxval = 0;
      if (i % 2) {
        int hn = n / 2;
        if (hn % 2 == 0) {
          maxval = 1 + (hn * hn / 2) * 4 + 4 * (hn / 2);
        } else {
          maxval = 1 + 4 * (hn * (hn + 1)) / 2;
        }
      } else {
        int hn = n / 2;
        if (hn % 2 == 0) {
          maxval = (hn * hn / 2) * 4 + 2 * (hn / 2) * 2;
        } else {
          maxval = 4 * (hn * (hn + 1)) / 2;
        }
      }
      if (maxval >= i) {
        ans[i] = min(ans[i], n);
        break;
      }
    }
  }
  ans[1] = 1;
  ans[2] = 3;
  ans[3] = 5;
  ans[4] = 3;
  ans[5] = 3;
  {
    int n;
    scanf("%d", &n);
    printf("%d\n", ans[n]);
  }
  return 0;
}

### Prompt

Please formulate 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 ans[128];
int main() {
  for (int i = 1; i <= 100; i++) {
    ans[i] = 10000;
    if (i % 4 == 0) {
      for (int n = 4;; n += 2) {
        int hn = n / 2 - 1;
        int maxval = 0;
        if (hn % 2) {
          maxval = 4 * ((hn * hn + 1) / 2);
        } else {
          maxval = 4 * (hn * hn / 2);
        }
        if (maxval >= i) {
          ans[i] = min(ans[i], n);
          break;
        }
      }
    }
    for (int n = 5;; n += 2) {
      int maxval = 0;
      if (i % 2) {
        int hn = n / 2;
        if (hn % 2 == 0) {
          maxval = 1 + (hn * hn / 2) * 4 + 4 * (hn / 2);
        } else {
          maxval = 1 + 4 * (hn * (hn + 1)) / 2;
        }
      } else {
        int hn = n / 2;
        if (hn % 2 == 0) {
          maxval = (hn * hn / 2) * 4 + 2 * (hn / 2) * 2;
        } else {
          maxval = 4 * (hn * (hn + 1)) / 2;
        }
      }
      if (maxval >= i) {
        ans[i] = min(ans[i], n);
        break;
      }
    }
  }
  ans[1] = 1;
  ans[2] = 3;
  ans[3] = 5;
  ans[4] = 3;
  ans[5] = 3;
  {
    int n;
    scanf("%d", &n);
    printf("%d\n", ans[n]);
  }
  return 0;
}

```