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 <algorithm>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;
 
int n,flag[100100][20],ans;
struct edge
{
    int v;
    edge *nxt;
} pool[200100],*tp=pool,*fst[100100];
 
void dfs(int x,int fa)
{
    for (edge *i=fst[x]; i; i=i->nxt)
        if (i->v!=fa)
            dfs(i->v,x);
    int w=0;
    repu(i,0,ans)
        if (flag[x][i]>1)
            w=i+1;
    for (; flag[x][w]; ++w);
    ans=max(ans,w),++flag[fa][w];
    repu(i,w+1,ans)
        if (flag[x][i])
            ++flag[fa][i];
}
 
int main()
{
    scanf("%d",&n);
    repu(i,1,n-1)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        *tp=(edge){v,fst[u]},fst[u]=tp++;
        *tp=(edge){u,fst[v]},fst[v]=tp++;
    }
    dfs(1,0);
    printf("%d\n",ans);
    return 0;
}