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 <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 1e5 + 5, M = N << 1 ;
using namespace std ;

int n, e, ter[M], nxt[M], lnk[N], f[N] ;

void add(int x, int y) {
	ter[++ e] = y, nxt[e] = lnk[x], lnk[x] = e ;
}

inline int count_max(int x) {
	int ret = - 1 ;
	rep(i, 0, 20) if ((x >> i) & 1) ret = i ;
	return ret ;
}

void dfs(int p, int fa) {
	int las = 0, must = 0 ;
	for (int i = lnk[p]; i; i = nxt[i]) if (ter[i] != fa) {
		dfs(ter[i], p) ;
		must |= las & f[ter[i]] ;
		las |= f[ter[i]] ;
	}
	int mx = count_max(must) + 1 ;
	for ( ; (1 << mx) & las ; ++ mx) ;
	f[p] = (las | (1 << mx)) >> mx << mx ;
}

int main() {
	scanf("%d", &n) ;
	int x, y ;
	rep(i, 1, n - 1) {
		scanf("%d%d", &x, &y) ;
		add(x, y), add(y, x) ;
	}
	dfs(1, 0) ;
	printf("%d\n", count_max(f[1])) ;
	return 0 ; 
}