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 <bits/stdc++.h>

using std::max;
using std::min;
const int N=1e5+10;

struct Edge{ int to,next; }way[N<<1];
int n,dep[N],num[N],tot,ans,val[N],cnt[N][21];

void Build(int a,int b) { way[++tot]=(Edge){b,num[a]}; num[a]=tot; }

void Init() {
	scanf("%d",&n);
	int a,b;
	for (int i=2;i<=n;++i) {
		scanf("%d%d",&a,&b);
		Build(a,b);
		Build(b,a);
	}
}

void Dfs(int x,int fa) {
	for (int i=num[x];i;i=way[i].next) {
		int v=way[i].to;
		if (v==fa) continue;
		Dfs(v,x);
		for (int i=val[v];i<=20;++i) cnt[x][i]+=cnt[v][i];
	}
	int value=0;
	for (int i=0;i<=19;++i) if (cnt[x][i]>=2) value=i+1;
	while (cnt[x][value]) ++value;
	val[x]=value; cnt[x][value]++;
	ans=max(ans,value);
}

void Solve() {
	Dfs(1,0);
	printf("%d\n",ans);
}

int main() {
	Init();
	Solve();
	return 0;
}