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 <bits/stdc++.h>
using namespace std;
#define MAXN 262144
#define rint register int
inline int read(){int r;int s=0,c;for(;!isdigit(c=getchar());s=c);for(r=c^48;isdigit(c=getchar());(r*=10)+=c^48);return s^45?r:-r;}
int c[MAXN+5], h[MAXN+5], n, _; vector<int> e[MAXN+5];
int DP(int p, int fa){int U = 0, S = 0; for(auto v:e[p]) if(v^fa) _ = DP(v,p), S |= _&U, U |= _; for(_ = h[S]; U&1<<_; ++_); return U>>_<<_|1<<_;}
int main()
{
	n = read(); for(rint i = 2, a, b; i <= n; a = read(), b = read(), e[a].push_back(b), e[b].push_back(a), i++);
	c[1] = 1; for(rint i = 2; i <= MAXN; c[i] = -~c[i-(i&-i)], h[i] = -~h[i>>1], i++); return !printf("%d\n",h[DP(1,0)]);
}