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 <algorithm>
using namespace std;

const int maxn=1e5+5;

int n,tot,ans;
int bit[maxn];
int now[maxn],pre[maxn*2],son[maxn*2];

int read() {
	int x=0,f=1;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
	return x*f;
}

void add(int a,int b) {
	pre[++tot]=now[a];
	now[a]=tot,son[tot]=b;
}

void dfs(int fa,int u) {
	int limit=0;
	for(int p=now[u],v=son[p];p;p=pre[p],v=son[p])
		if(v!=fa) {
			dfs(u,v);
			limit|=bit[u]&bit[v];
			bit[u]|=bit[v];
		}
	int mx=20;
	while(mx>=0&&(!(limit&(1<<mx))))mx--;
	mx++;
	while(bit[u]&(1<<mx))mx++;
	bit[u]|=1<<mx;
	ans=max(ans,mx),bit[u]=bit[u]>>mx<<mx;
}

int main() {
	n=read();
	for(int i=1;i<n;i++) {
		int a=read(),b=read();
		add(a,b),add(b,a);
	}
	dfs(0,1);
	printf("%d\n",ans);
	return 0;
}
