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>

const int N=100005;

int fst[N],v[N],mask[N],tot=0;

struct Edge{
	int to,nxt;
}e[N*2];

void addedge(int x,int y){
	e[++tot].to=y;e[tot].nxt=fst[x];fst[x]=tot;
}

void dfs(int x,int father){
	mask[x]=0;v[x]=0;
	for (int i=fst[x];i;i=e[i].nxt){
		int y=e[i].to;
		if (y!=father){
			dfs(y,x);
			int s=mask[x]&mask[y],j=0;
			mask[x]|=mask[y];
			for (;s;s>>=1) j++;
			if (j>v[x]) v[x]=j;
			for (;(mask[x]>>v[x])&1;v[x]++);
			for (j=0;j<v[x];j++) mask[x]^=mask[x]&(1<<j);
		}
	}
	mask[x]|=1<<v[x];
}

int main(){
	int n;
	scanf("%d\n",&n);
	for (int i=1;i<n;i++){
		int x,y;
		scanf("%d%d\n",&x,&y);
		addedge(x,y);
		addedge(y,x);
	}
	dfs(1,0);
	int ans=0;
	for (int i=1;i<=n;i++)
		if (v[i]>ans) ans=v[i];
	printf("%d\n",ans);
}