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>
using namespace std;
const int N=100100;
int i,j,k,n,En,ch,ff,ans,x,y;
int h[N],f[N];
struct edge { int s,n;} E[N<<1];
void R(int &x) {
	ff=x=0;ch=getchar();
	while (ch<'0' || '9'<ch) { if (ch=='-') ff=1;ch=getchar();}
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
}
void E_add(int x,int y) {
	E[++En].s=y;E[En].n=h[x];h[x]=En;
	E[++En].s=x;E[En].n=h[y];h[y]=En;
}
void dfs(int x,int F) {
	int t=0,tt=0;
	for (int k=h[x];k;k=E[k].n) if (E[k].s!=F) {
		dfs(E[k].s,x);
		tt|=f[E[k].s]&t;
		t|=f[E[k].s];
	}
	int i=0;
	while (((1<<i)<=tt) || ((1<<i)&t)) i++;
	ans=max(ans,i);
	f[x]=(t>>i<<i)|(1<<i);
}
int main() {
	R(n);
	for (i=1;i<n;i++) {
		R(x);R(y);
		E_add(x,y);
	}
	dfs(1,0);
	printf("%d\n",ans);
}