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 <cstring>
#include <algorithm>
#include <cstdio>

using std::max_element;
const int N = 100010;
const int M = 30;
int _w;

namespace Tree {
	int head[N], nxt[N*2], to[N*2], eid;
	void init() {
		eid = 0;
		memset(head, -1, sizeof head);
	}
	void link( int u, int v ) {
		to[eid] = v, nxt[eid] = head[u], head[u] = eid++;
		to[eid] = u, nxt[eid] = head[v], head[v] = eid++;
	}
}

int n, f[N][M], g[N];

void dfs( int u, int fa ) {
	using namespace Tree;
	for( int i = head[u]; ~i; i = nxt[i] ) {
		int v = to[i];
		if( v == fa ) continue;
		dfs(v, u);
		for( int j = 0; j < M; ++j )
			f[u][j] += f[v][j];
	}
	int i;
	for( i = M-1; i; --i )
		if( f[u][i] >= 2 )
			break;
	while( f[u][i] ) ++i;
	g[u] = i, f[u][i] = 1;
	for( int j = 0; j < i; ++j )
		f[u][j] = 0;
}

int main() {
	_w = scanf( "%d", &n );
	Tree::init();
	for( int i = 1; i < n; ++i ) {
		int u, v;
		_w = scanf( "%d%d", &u, &v );
		Tree::link(u, v);
	}
	dfs(1, 0);
	printf( "%d\n", *max_element(g+1, g+n+1) );
	return 0;
}
