#include <bits/stdc++.h>
using namespace std;
int a[1001][1001];
bool can(int x, int n) {
  int kek = 1;
  if (n & 1) {
    if (x % 2 == 0) return false;
    a[x / 2 + 1][x / 2 + 1] = 1;
    --n;
    kek = 0;
  }
  for (int i = 1; i <= x; i++)
    for (int j = 1; j <= x; j++) a[i][j] = 0;
  for (int i = 1; i <= x / 2 - (x % 2 == 0); i++)
    for (int j = 1; j <= x / 2 - (x % 2 == 0); j++)
      if (n >= 4 && !a[i - 1][j] && !a[i][j - 1]) {
        n -= 4;
        a[i][j] = 1;
      }
  if (x & 1) {
    for (int i = 1; i < x / 2 + kek; i++)
      if (!a[x / 2][i] && !a[x / 2 + 1][i - 1] && n >= 2) {
        n -= 2;
        a[x / 2 + 1][i] = 1;
      }
    for (int i = 1; i < x / 2 + kek; i++)
      if (!a[i][x / 2] && !a[i - 1][x / 2 + 1] && n >= 2) {
        n -= 2;
        a[i][x / 2 + 1] = 1;
      }
  }
  return n == 0;
}
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << 1 << "\n";
    return 0;
  }
  for (int i = 3; i <= 1000; i++)
    if (can(i, n)) {
      cout << i << "\n";
      return 0;
    }
}

### Prompt

Your challenge is to write 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 a[1001][1001];
bool can(int x, int n) {
  int kek = 1;
  if (n & 1) {
    if (x % 2 == 0) return false;
    a[x / 2 + 1][x / 2 + 1] = 1;
    --n;
    kek = 0;
  }
  for (int i = 1; i <= x; i++)
    for (int j = 1; j <= x; j++) a[i][j] = 0;
  for (int i = 1; i <= x / 2 - (x % 2 == 0); i++)
    for (int j = 1; j <= x / 2 - (x % 2 == 0); j++)
      if (n >= 4 && !a[i - 1][j] && !a[i][j - 1]) {
        n -= 4;
        a[i][j] = 1;
      }
  if (x & 1) {
    for (int i = 1; i < x / 2 + kek; i++)
      if (!a[x / 2][i] && !a[x / 2 + 1][i - 1] && n >= 2) {
        n -= 2;
        a[x / 2 + 1][i] = 1;
      }
    for (int i = 1; i < x / 2 + kek; i++)
      if (!a[i][x / 2] && !a[i - 1][x / 2 + 1] && n >= 2) {
        n -= 2;
        a[i][x / 2 + 1] = 1;
      }
  }
  return n == 0;
}
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << 1 << "\n";
    return 0;
  }
  for (int i = 3; i <= 1000; i++)
    if (can(i, n)) {
      cout << i << "\n";
      return 0;
    }
}

```