#include <bits/stdc++.h>
using namespace std;
int x;
int f(int x) { return (x * x + 1) / 2; }
int i;
int main() {
  cin >> x;
  for (i = 1;; i++) {
    if (i % 2 == 0) {
      if (x % 4 == 0 && f(i / 2 - 1) * 4 >= x) {
        cout << i;
        return 0;
      }
    } else {
      int cnt_indep = 1;
      int cnt_twos = (i % 4 == 1 ? (i - 1) : (i - 3));
      int cnt_all = f(i) - cnt_indep - cnt_twos;
      for (int i = 0; i <= cnt_indep; i++) {
        for (int j = 0; j <= cnt_twos; j += 2) {
          for (int k = 0; k <= cnt_all; k += 4) {
            if (i + j + k == x) {
              cout << ::i;
              return 0;
            }
          }
        }
      }
      if (i % 4 == 3) {
        int cnt_indep = 0;
        int cnt_twos = i + 1;
        int cnt_all = 4 * (f(i / 2) - 1);
        for (int i = 0; i <= cnt_indep; i++) {
          for (int j = 0; j <= cnt_twos; j += 2) {
            for (int k = 0; k <= cnt_all; k += 4) {
              if (i + j + k == x) {
                cout << ::i;
                return 0;
              }
            }
          }
        }
      }
    }
  }
}

### Prompt

Your task is to create 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 x;
int f(int x) { return (x * x + 1) / 2; }
int i;
int main() {
  cin >> x;
  for (i = 1;; i++) {
    if (i % 2 == 0) {
      if (x % 4 == 0 && f(i / 2 - 1) * 4 >= x) {
        cout << i;
        return 0;
      }
    } else {
      int cnt_indep = 1;
      int cnt_twos = (i % 4 == 1 ? (i - 1) : (i - 3));
      int cnt_all = f(i) - cnt_indep - cnt_twos;
      for (int i = 0; i <= cnt_indep; i++) {
        for (int j = 0; j <= cnt_twos; j += 2) {
          for (int k = 0; k <= cnt_all; k += 4) {
            if (i + j + k == x) {
              cout << ::i;
              return 0;
            }
          }
        }
      }
      if (i % 4 == 3) {
        int cnt_indep = 0;
        int cnt_twos = i + 1;
        int cnt_all = 4 * (f(i / 2) - 1);
        for (int i = 0; i <= cnt_indep; i++) {
          for (int j = 0; j <= cnt_twos; j += 2) {
            for (int k = 0; k <= cnt_all; k += 4) {
              if (i + j + k == x) {
                cout << ::i;
                return 0;
              }
            }
          }
        }
      }
    }
  }
}

```