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<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
int n, D[101000];
vector<int>E[101000];
void DFS(int a, int pp){
    int i;
    int g=0,h=0,tp=0;
    for(i=0;i<E[a].size();i++){
        int x = E[a][i];
        if(x==pp)continue;
        DFS(x, a);
        h |= g&D[x];
        g |= D[x];
    }
    while(h){
        tp=(h&-h);
        h^=tp;
    }
    if(!tp)tp=1;
    while(tp&g)tp<<=1;
    D[a] = (tp|g)/tp*tp;
}
int main(){
    int i, a, b;
    scanf("%d",&n);
    for(i=1;i<n;i++){
        scanf("%d%d",&a,&b);
        E[a].push_back(b);
        E[b].push_back(a);
    }
    DFS(1,0);
    for(i=30;i>=0;i--)if((1<<i)&D[1])break;
    printf("%d\n",i);
}