#include <bits/stdc++.h>
using namespace std;
template <class T>
T f_min(T x, T y) {
  return x < y ? x : y;
}
template <class T>
T f_max(T x, T y) {
  return x > y ? x : y;
}
template <class T>
T f_abs(T x) {
  return x < 0 ? -x : x;
}
int num[100];
void pre() {
  int i, t;
  num[1] = 1;
  num[2] = 0;
  for (i = 3; i < 100; i++) {
    t = (i + 1) / 2;
    if (i & 1) {
      num[i] = i * i / 2 + 1;
    } else {
      num[i] = (((t - 1) * (t - 1) + 1) / 2) * 4;
    }
  }
}
int N;
void get_data() { scanf("%d", &N); }
int run() {
  if (N == 1) return 1;
  if (N == 2) return 3;
  if (N == 3) return 5;
  int i;
  for (i = 1; i < 100; i++) {
    if (i & 1) {
      if (num[i] >= N) break;
    } else {
      if (num[i] >= N && (num[i] - N) % 4 == 0) break;
    }
  }
  return i;
}
int main() {
  pre();
  get_data();
  printf("%d\n", run());
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <class T>
T f_min(T x, T y) {
  return x < y ? x : y;
}
template <class T>
T f_max(T x, T y) {
  return x > y ? x : y;
}
template <class T>
T f_abs(T x) {
  return x < 0 ? -x : x;
}
int num[100];
void pre() {
  int i, t;
  num[1] = 1;
  num[2] = 0;
  for (i = 3; i < 100; i++) {
    t = (i + 1) / 2;
    if (i & 1) {
      num[i] = i * i / 2 + 1;
    } else {
      num[i] = (((t - 1) * (t - 1) + 1) / 2) * 4;
    }
  }
}
int N;
void get_data() { scanf("%d", &N); }
int run() {
  if (N == 1) return 1;
  if (N == 2) return 3;
  if (N == 3) return 5;
  int i;
  for (i = 1; i < 100; i++) {
    if (i & 1) {
      if (num[i] >= N) break;
    } else {
      if (num[i] >= N && (num[i] - N) % 4 == 0) break;
    }
  }
  return i;
}
int main() {
  pre();
  get_data();
  printf("%d\n", run());
  return 0;
}

```