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;

typedef long long ll;

int N;
vector<int> G[100010];

ll dp[100010];

ll DFS(int v,int p)
{
    ll AND = 0,OR = 0;
    if(G[v].size() == 1 && G[v][0] == p)return dp[v] = 1;
    for(int i = 0; i < G[v].size(); i++)
    {
        int ne = G[v][i];
        if(ne == p)continue;
        ll now = DFS(ne,v);
        AND |= (now & OR);
        OR |= now;
    }
    int id = 0;
    while(AND >= (1 << id))id++;
    while((OR >> id) & 1)id++;
    return dp[v] = (OR + (1 << id)) & (~((1 << id) - 1));
}

int main()
{
    scanf("%d",&N);
    for(int i = 0; i < N - 1; i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
    }

    ll res = DFS(1,-1);

    for(int i = 0; res >> i; i++)
    {
        if(res >> i + 1 == 0)
        {
            printf("%d\n",i);
            return 0;
        }
    }
}
