We will recursively define uninity of a tree, as follows: (Uni is a Japanese word for sea urchins.)

* A tree consisting of one vertex is a tree of uninity 0.
* Suppose there are zero or more trees of uninity k, and a vertex v. If a vertex is selected from each tree of uninity k and connected to v with an edge, the resulting tree is a tree of uninity k+1.



It can be shown that a tree of uninity k is also a tree of uninity k+1,k+2,..., and so forth.

You are given a tree consisting of N vertices. The vertices of the tree are numbered 1 through N, and the i-th of the N-1 edges connects vertices a_i and b_i.

Find the minimum k such that the given tree is a tree of uninity k.

Constraints

* 2 ≦ N ≦ 10^5
* 1 ≦ a_i, b_i ≦ N(1 ≦ i ≦ N-1)
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

Print the minimum k such that the given tree is a tree of uninity k.

Examples

Input

7
1 2
2 3
2 4
4 6
6 7
7 5


Output

2


Input

12
1 2
2 3
2 4
4 5
5 6
6 7
7 8
5 9
9 10
10 11
11 12


Output

3
import sys
range = xrange

def decomp(coupl, root = 0):
    n = len(coupl)
    visible_labels = [0] * n
    
    bfs = [root]
    for node in bfs:
        for nei in coupl[node]:
            coupl[nei].remove(node)
        bfs += coupl[node]

    for node in reversed(bfs):
        seen = seen_twice = 0
        for nei in coupl[node]:
            seen_twice |= seen & visible_labels[nei]
            seen |= visible_labels[nei]
        tmp = ~seen & -(1 << seen_twice.bit_length()) 
        label = tmp & -tmp
        visible_labels[node] = (label | seen) & -label
    
    return [(seen & -seen).bit_length() - 1 for seen in visible_labels]

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0

n = inp[ii]; ii += 1
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
    u = inp[ii] - 1; ii += 1
    v = inp[ii] - 1; ii += 1
    coupl[u].append(v)
    coupl[v].append(u)

print max(decomp(coupl))
