Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

2

A tree of uninity 1 consisting of vertices 1, 2, 3 and 4 can be constructed from the following: a tree of uninity 0 consisting of vertex 1, a tree of uninity 0 consisting of vertex 3, a tree of uninity 0 consisting of vertex 4, and vertex 2.

A tree of uninity 1 consisting of vertices 5 and 7 can be constructed from the following: a tree of uninity 1 consisting of vertex 5, and vertex 7.

A tree of uninity 2 consisting of vertices 1, 2, 3, 4, 5, 6 and 7 can be constructed from the following: a tree of uninity 1 consisting of vertex 1, 2, 3 and 4, a tree of uninity 1 consisting of vertex 5 and 7, and vertex 6.

Sample Input 2

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

Sample Output 2

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<algorithm>
using namespace std;
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define MN 100000
struct edge{int nx,t;}e[MN*2+5];
int h[MN+5],en,f[MN+5];
inline void ins(int x,int y)
{
	e[++en]=(edge){h[x],y};h[x]=en;
	e[++en]=(edge){h[y],x};h[y]=en;
}
int cal(int x){for(int i=17;i;--i)if(x&(1<<i))return i;return -1;}
void dfs(int x,int fa)
{
	int mx=0;
	for(int i=h[x];i;i=e[i].nx)if(e[i].t!=fa)
	{
		dfs(e[i].t,x);
		mx=max(mx,cal(f[x]&f[e[i].t])+1);
		f[x]|=f[e[i].t];
	}
	for(int i=0;;++i)if(i>=mx&&!(f[x]&(1<<i)))
		{f[x]=(f[x]|(1<<i))&((-1)<<i);return;}
}
int main()
{
	int n=read(),i;
	for(i=1;i<n;++i)ins(read(),read());
	dfs(1,0);
	printf("%d",cal(f[1]));
}