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
#include <cstdio>
#include <set>

int n,head[100010],nxt[200010],b[200010],k,ans;
void push(int s,int t){
	nxt[++k]=head[s];
	head[s]=k;
	b[k]=t;
}
std::set<int>set[100010];
void dfs(int x,int f){
	int mx=0;
	for(int i=head[x];i;i=nxt[i])
		if(b[i]!=f){
			dfs(b[i],x);
			if(set[b[i]].size()>mx)mx=b[i];
		}
	set[x]=set[mx];
	int v=0;
	for(int i=head[x];i;i=nxt[i])
		if(b[i]!=mx){
			for(auto it=set[b[i]].begin();it!=set[b[i]].end();++it)
				if(set[x].count(*it)){
					set[x].erase(*it);
					v=std::max(v,*it+1);
				}
				else if(v<=*it){
					set[x].insert(*it);
				}
		}
	while(set[x].count(v))++v;
	while(!set[x].empty()&&*set[x].begin()<v)set[x].erase(*set[x].begin());
	set[x].insert(v);
//	printf("%d %d\n",x,v);
	ans=std::max(ans,v);
}
int main(){
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++)scanf("%d%d",&u,&v),push(u,v),push(v,u);
	dfs(1,0);
	printf("%d\n",ans);
}