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
//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define F first
#define S second
#define _sz(x) (int)x.size()

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 1e5 + 20 ;
int n , val[N] ;
vector <int> g[N] ;

inline int best (int x) {
	for (int i = 20 ; i >= 0 ; i --)
		if (x >> i & 1) return i ;

	return -1 ;
}

void dfs (int v , int par = -1) {
	int mx = -1 ;

	for (int u : g[v]) {
		if (u == par) continue ;

		dfs(u , v) ;

		mx = max(mx , best(val[v] & val[u])) ;

		val[v] |= val[u] ;
	}

	for (int i = 0 ; i <= 20 ; i ++) {
		if (i <= mx || (val[v] & (1 << i))) {
			val[v] = ((val[v] | (1 << i)) ^ (1 << i)) ;
			continue ;
		}

		val[v] = (val[v] | (1 << i)) ;
		break ;
	}
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	cin >> n ;

	for (int i = 0 , u , v ; i < n - 1 ; i ++) {
		cin >> u >> v ;
		u -- , v -- ;
		g[u].push_back(v) ;
		g[v].push_back(u) ;
	}

	dfs(0) ;

	cout << best(val[0]) << '\n' ;
}
