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>
"""Z = [1<<62]*102

M = []

def magic(i, j, n, s):
	if s > 100:
		return
	Z[s] = min(Z[s], n)
	if j > (n+1)/2:
		return magic(i+1, 1, n, s)
	if i > (n+1)/2:
		return
	if M[i-1][j] == 0 and M[i][j-1] == 0:
		M[i][j] = 1
		t = 4
		if i == (n+1)/2:
			t /= 2
		if j == (n+1)/2:
			t /= 2
		magic(i, j+1, n, s+t)
		M[i][j] = 0
	magic(i, j+1, n, s)

for n in range(1, 12):
	M = [[0]*(n+2) for _ in range(n+2)]
	magic(1, 1, n, 0)
	print n

print Z
"""

Z = [1, 1, 3, 5, 3, 3]
n = 8
for i in range(5, 16, 2):
	Z += [i]*n
	n += 4

x = input()
print Z[x]
