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>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N;
vint G[111111];

int dfs(int v,int p){
    vint lis;
    for(auto u:G[v]){
        if(u==p)continue;
        lis.pb(dfs(u,v));
    }

    int k=0;
    int ret=0;
    for(auto x:lis)ret|=x;
    rep(i,20){
        int cnt=0;
        for(auto x:lis)if(x>>i&1)cnt++;
        if(cnt>1)k=i+1;
    }
    while(ret>>k&1)k++;
    rep(i,k)if(ret>>i&1)ret^=1<<i;
    ret^=1<<k;
    return ret;
}

signed main(){
    cin>>N;
    rep(i,N-1){
        int a,b;
        cin>>a>>b;
        a--;b--;
        G[a].pb(b);G[b].pb(a);
    }
    int tmp=dfs(0,-1);
    int k=20;
    while(~tmp>>k&1)k--;
    cout<<k<<endl;
    return 0;
}
