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>
#include <bits/stdc++.h>
using namespace std;
int ans[110];
void update(int a, int b, int c, int n) {
  for (int i = 0; i <= a; i++)
    for (int j = 0; j <= b; j++)
      for (int k = 0; k <= c; k++) {
        int m = i * 4 + j * 2 + k;
        if (m < 110) ans[m] = min(ans[m], n);
      }
}
void init() {
  int i, j;
  for (i = 1; i < 110; i++) ans[i] = 1000000000;
  ans[1] = 1;
  for (i = 1; i < 110; i++) {
    if (i % 2) {
      update(i * i / 2, (i / 2 + 1) * 2, 0, i * 2 + 1);
      update(i * i / 2 + 1, (i / 2) * 2, 1, i * 2 + 1);
    } else {
      update(i * i / 2, (i / 2) * 2, 1, i * 2 + 1);
    }
  }
  for (i = 1; i < 110; i++) {
    if (i % 2)
      update((i - 1) * (i - 1) / 2, 0, 0, i * 2);
    else
      update((i - 1) * (i - 1) / 2 + 1, 0, 0, i * 2);
  }
}
int main() {
  init();
  int n;
  scanf("%d", &n);
  cout << ans[n] << endl;
  return 0;
}
