#include <bits/stdc++.h>
using namespace std;
int solve0(int x) {
  int sum = 0;
  for (int n = 3, k = 3; true; n += 2, k++) {
    sum += (k - 2) * 4;
    if (sum >= x) return cout << n << endl, 0;
  }
}
int solve1(int x) {
  int sum = 1;
  for (int n = 3, k = 3; true; n += 2, k++) {
    sum += (k - 2) * 4;
    if (sum >= x) return cout << n << endl, 0;
  }
}
int solve2(int x) {
  int sum = 2;
  if (x == 2) return cout << 3 << endl, 0;
  for (int n = 3, k = 3; true; n += 2, k++) {
    sum += (k - 1) * 4;
    if (sum >= x) return cout << n + 2 << endl, 0;
  }
}
int solve3(int x) {
  if (x == 3 || x == 7 || x == 11) return cout << 5 << endl, 0;
  int sum = 11;
  for (int n = 7, k = 7; true; n += 2, k++) {
    sum += (k - 4) * 4;
    if (sum >= x) return cout << n << endl, 0;
  }
}
int main() {
  int x;
  cin >> x;
  if (x == 1) return cout << 1 << endl, 0;
  if (x % 4 == 0) solve0(x);
  if (x % 4 == 1) solve1(x);
  if (x % 4 == 2) solve2(x);
  if (x % 4 == 3) solve3(x);
  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 solve0(int x) {
  int sum = 0;
  for (int n = 3, k = 3; true; n += 2, k++) {
    sum += (k - 2) * 4;
    if (sum >= x) return cout << n << endl, 0;
  }
}
int solve1(int x) {
  int sum = 1;
  for (int n = 3, k = 3; true; n += 2, k++) {
    sum += (k - 2) * 4;
    if (sum >= x) return cout << n << endl, 0;
  }
}
int solve2(int x) {
  int sum = 2;
  if (x == 2) return cout << 3 << endl, 0;
  for (int n = 3, k = 3; true; n += 2, k++) {
    sum += (k - 1) * 4;
    if (sum >= x) return cout << n + 2 << endl, 0;
  }
}
int solve3(int x) {
  if (x == 3 || x == 7 || x == 11) return cout << 5 << endl, 0;
  int sum = 11;
  for (int n = 7, k = 7; true; n += 2, k++) {
    sum += (k - 4) * 4;
    if (sum >= x) return cout << n << endl, 0;
  }
}
int main() {
  int x;
  cin >> x;
  if (x == 1) return cout << 1 << endl, 0;
  if (x % 4 == 0) solve0(x);
  if (x % 4 == 1) solve1(x);
  if (x % 4 == 2) solve2(x);
  if (x % 4 == 3) solve3(x);
  return 0;
}

```