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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 100005;

int ehead[maxn],ecnt;
struct edge {
	int u,v,next;
}edg[maxn<<1];
void add(int u,int v) {
	edg[++ecnt]=(edge){u,v,ehead[u]};
	ehead[u]=ecnt;
	edg[++ecnt]=(edge){v,u,ehead[v]};
	ehead[v]=ecnt;
}
int n,bit[maxn],f[maxn],ans;
void dfs(int u,int fa)
{
	int conf=0;
	for (int v,j=ehead[u];j;j=edg[j].next)
	if ((v=edg[j].v)!=fa) {
		dfs(v,u);conf|=bit[u]&bit[v];bit[u]|=bit[v];
	}
	while ((conf>>f[u])||(bit[u]>>f[u]&1)) ++f[u];
	bit[u]=(bit[u]>>f[u]<<f[u])|(1<<f[u]);
	ans=max(ans,f[u]);
}
int main()
{
	#ifdef Amberframe
		freopen("agc009d.in","r",stdin);
		freopen("agc009d.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int u,v,i=1;i<n;i++)
		scanf("%d %d",&u,&v),add(u,v);
	dfs(1,0);printf("%d",ans);
	return 0;
}