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 <cstring>
#include <algorithm>
#include <cstdlib>
#include <queue>
using namespace std;
#define N 100050
int head[N],to[N<<1],nxt[N<<1],cnt,n,tmp[22],f[N][22],ans;
inline void add(int u,int v) {
	to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt;
}
void merge(int *a,int *b) {
	int i;
	for(i=0;i<=20;i++) a[i]+=b[i];
}
void dfs(int x,int y) {
	int i;
	for(i=head[x];i;i=nxt[i]) if(to[i]!=y) dfs(to[i],x);
	memset(tmp,0,sizeof(tmp));
	for(i=head[x];i;i=nxt[i]) if(to[i]!=y) merge(tmp,f[to[i]]);
	int k=0;
	for(i=20;i>=0;i--) {
		if(tmp[i]>1) {
			k=i+1; break;
		}
	}
	for(;tmp[k];k++) ;
	memcpy(f[x]+k,tmp+k,sizeof(int)*(20-k));
	f[x][k]++;
	ans=max(ans,k);
}
int main() {
	scanf("%d",&n);
	int i,x,y;
	for(i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	dfs(1,0); printf("%d\n",ans);
}
