#include <bits/stdc++.h>
int n;
int d[16];
int solution(int x) {
  int i;
  int sol;
  for (i = 0; i < 16; i++) {
    if (x <= d[i]) {
      if (i % 2 == 0) {
        if (((x - d[i]) % 4) == 0) sol = i;
      } else {
        sol = i;
        break;
      }
    }
  }
  return sol;
}
int main() {
  int x;
  scanf("%d", &x);
  d[0] = 0;
  d[1] = 1;
  d[2] = 0;
  d[3] = 5;
  d[4] = d[1] * 4;
  d[5] = 13;
  d[6] = 2 * 4;
  d[7] = 25;
  d[8] = d[3] * 4;
  d[9] = 41;
  d[10] = d[4] * 4;
  d[11] = 61;
  d[12] = d[5] * 4;
  d[13] = 85;
  d[14] = d[6] * 4;
  d[15] = 113;
  if (x == 3)
    printf("5\n");
  else
    printf("%d\n", solution(x));
  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>
int n;
int d[16];
int solution(int x) {
  int i;
  int sol;
  for (i = 0; i < 16; i++) {
    if (x <= d[i]) {
      if (i % 2 == 0) {
        if (((x - d[i]) % 4) == 0) sol = i;
      } else {
        sol = i;
        break;
      }
    }
  }
  return sol;
}
int main() {
  int x;
  scanf("%d", &x);
  d[0] = 0;
  d[1] = 1;
  d[2] = 0;
  d[3] = 5;
  d[4] = d[1] * 4;
  d[5] = 13;
  d[6] = 2 * 4;
  d[7] = 25;
  d[8] = d[3] * 4;
  d[9] = 41;
  d[10] = d[4] * 4;
  d[11] = 61;
  d[12] = d[5] * 4;
  d[13] = 85;
  d[14] = d[6] * 4;
  d[15] = 113;
  if (x == 3)
    printf("5\n");
  else
    printf("%d\n", solution(x));
  return 0;
}

```