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<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int n;
std::vector<int> to[maxn];
int cnt[maxn][20],id[maxn];
void dfs(int x,int f=0){
	for(auto i:to[x])if(i!=f){
		dfs(i,x);
		rep(j,0,19)cnt[x][j]+=cnt[i][j];
	}
	int&min=id[x]=0;
	rep(j,0,19)if(cnt[x][j]>=2)min=j+1;
	for(;min<20&&cnt[x][min];)++min;
	rep(j,0,min-1)cnt[x][j]=0;
	cnt[x][min]=1;
}
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	rep(i, 1, n - 1) {
		int x, y; cin >> x >> y;
		to[x].push_back(y);
		to[y].push_back(x);
	}
	dfs(1), cout<<*std::max_element(id+1,id+n+1)<<'\n';;
}
