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 <cstdio>
#include <vector>

std::vector<int> edges[100005];

int uninity[100005];

int msb[1<<20];

void dfs(int node,int parent){
  int once=0,twice=0;
  for(int child:edges[node]){
    if(child==parent) continue;
    dfs(child,node);
    twice|=(uninity[child]&once);
    once|=uninity[child];
  }
  uninity[node]=(once&~((1<<msb[twice])-1))+(1<<msb[twice]);
  //printf("uninity[%d]=%d\n",node,uninity[node]);
}

int main(){
  for(int i=1;i<(1<<20);i++){
    msb[i]=msb[i>>1]+1;
  }
  int N;
  scanf("%d",&N);
  for(int i=0;i<N-1;i++){
    int A,B;
    scanf("%d %d",&A,&B);
    A--,B--;
    edges[A].push_back(B);
    edges[B].push_back(A);
  }
  dfs(0,0);
  printf("%d\n",msb[uninity[0]]-1);
  return 0;
}
