#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = 10 * x + c - '0';
    c = getchar();
  }
  x *= f;
}
int main() {
  int n;
  read(n);
  if (n == 1) {
    printf("1");
    return 0;
  }
  if (n == 2 || n == 4 || n == 5) {
    printf("3");
    return 0;
  }
  for (int i = 5; i <= 10000; i++) {
    if (i & 1) {
      int l = 3, r = (i * i + 1) / 2;
      if (n >= l && n <= r) {
        printf("%d", i);
        return 0;
      }
    } else if (n % 4 == 0) {
      int x = n, mx = (i - 2) / 2;
      mx = (mx * mx + 1) / 2;
      if (mx * 4 >= x) {
        printf("%d", i);
        return 0;
      }
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = 10 * x + c - '0';
    c = getchar();
  }
  x *= f;
}
int main() {
  int n;
  read(n);
  if (n == 1) {
    printf("1");
    return 0;
  }
  if (n == 2 || n == 4 || n == 5) {
    printf("3");
    return 0;
  }
  for (int i = 5; i <= 10000; i++) {
    if (i & 1) {
      int l = 3, r = (i * i + 1) / 2;
      if (n >= l && n <= r) {
        printf("%d", i);
        return 0;
      }
    } else if (n % 4 == 0) {
      int x = n, mx = (i - 2) / 2;
      mx = (mx * mx + 1) / 2;
      if (mx * 4 >= x) {
        printf("%d", i);
        return 0;
      }
    }
  }
  return 0;
}

```