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
/* Bismillahir Rahmanir Rahim */

#include <bits/stdc++.h>

#define rep(i, n)	for(int i=0;i<n;i++)
#define repn(i, n)	for(int i=1;i<=n;i++)
#define set(i, n)	memset(i, n, sizeof(i))

#define pb	push_back
#define f	first
#define s	second

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

const int N = 100007;

int n, fx;
vector<int>edge[N];

int dfs(int at, int past){
	int mask = 0, cnt[25];
	set(cnt, 0);
	rep(i, edge[at].size()){
		int u = edge[at][i];
		if(u == past) continue;
		int fask = dfs(u, at);
		rep(j, 20){
			if(fask & (1 << j)) cnt[j]++;
		}
		mask |= fask;
	}
	int s = 0;
	for(int i=20;i>=0;i--){
		 if(cnt[i] < 2) continue;
		 s = i + 1;
		 break;
	} 
	for(int i=s;i<=20;i++){
		if(cnt[i] == 0){
			for(int j=0;j<i;j++) if(mask & (1 << j)) mask ^= (1 << j);
			mask |= (1 << i);
			return mask;
		}
	}
}

int main(){
	int x, y;
	scanf("%d", &n);
	repn(i, n-1){
		scanf("%d %d", &x, &y);
		edge[x].pb(y);
		edge[y].pb(x);
	}
	int ret = dfs(1, 1);
	for(int i=20;i>=0;i--){
		if(ret & (1 << i)){
			printf("%d\n", i);
			return 0;
		}
	}
	return 0;
}
