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 fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=100000+10;
int h[maxn],go[maxn*2],nxt[maxn*2],bit[maxn],f[maxn];
int i,j,k,l,t,n,m,tot,top,ans;
void add(int x,int y){
    go[++tot]=y;
    nxt[tot]=h[x];
    h[x]=tot;
}
void dfs(int x,int y){
    bit[x]=f[x]=0;
    int t=h[x],l=0;
    while (t){
        if (go[t]!=y){
            dfs(go[t],x);
            l=max(l,(bit[x]&bit[go[t]]));
            bit[x]|=bit[go[t]];
        }
        t=nxt[t];
    }
    if (!bit[x]) bit[x]=1;
    else{
        while ((1<<f[x])<=l||(1<<f[x])&bit[x]) f[x]++;
        bit[x]=(bit[x]>>f[x]<<f[x])|(1<<f[x]);
    }
    ans=max(ans,f[x]);
}
int main(){
    scanf("%d",&n);
    fo(i,1,n-1){
        scanf("%d%d",&j,&k);
        add(j,k);add(k,j);
    }
    dfs(1,0);
    printf("%d\n",ans);
}