#include <bits/stdc++.h>
using namespace std;
constexpr int INF = 1e9 + 1;
constexpr long long LLINF = 1e18 + 1;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
vector<pair<int, int> > sol;
void init() {
  sol.emplace_back(0, 0);
  int odd = 1, add4 = 1;
  pair<int, int> pre;
  for (int i = 1; i < 10; i++) {
    pre = sol.back();
    if (odd) {
      pre.second += add4;
      sol.emplace_back(pre);
    } else {
      pre.first += 2;
      pre.second += add4;
      add4 += 2;
      sol.emplace_back(pre);
    }
    odd = !odd;
  }
}
int get_ans(int x) {
  int t, i;
  for (i = 0; i < 10; i++) {
    t = x;
    t -= min(t / 4, sol[i].second) * 4;
    t -= min(t / 2, sol[i].first) * 2;
    t -= min(t, 1);
    if (!t) break;
  }
  return 2 * i + 1;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int x;
  init();
  ;
  while (cin >> x) {
    if (x == 2) {
      printf("3\n");
      continue;
    }
    printf("%d\n", get_ans(x));
  }
}

### 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;
constexpr int INF = 1e9 + 1;
constexpr long long LLINF = 1e18 + 1;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
vector<pair<int, int> > sol;
void init() {
  sol.emplace_back(0, 0);
  int odd = 1, add4 = 1;
  pair<int, int> pre;
  for (int i = 1; i < 10; i++) {
    pre = sol.back();
    if (odd) {
      pre.second += add4;
      sol.emplace_back(pre);
    } else {
      pre.first += 2;
      pre.second += add4;
      add4 += 2;
      sol.emplace_back(pre);
    }
    odd = !odd;
  }
}
int get_ans(int x) {
  int t, i;
  for (i = 0; i < 10; i++) {
    t = x;
    t -= min(t / 4, sol[i].second) * 4;
    t -= min(t / 2, sol[i].first) * 2;
    t -= min(t, 1);
    if (!t) break;
  }
  return 2 * i + 1;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int x;
  init();
  ;
  while (cin >> x) {
    if (x == 2) {
      printf("3\n");
      continue;
    }
    printf("%d\n", get_ans(x));
  }
}

```