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;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

const int MAXN = 100005, B = 20;
vector<int> gra[MAXN];
int f[MAXN][B], n, ans;

void dfs(int u, int fa) {
	for (int v : gra[u]) if (v != fa) {
		dfs(v, u);
		for (int i = 0; i < B; i++)
			f[u][i] += f[v][i];
	}
	int p = 0;
	for (int i = B - 1; ~i; i--)
		if (f[u][i] > 1) { p = i; break; }
	while (f[u][p]) ++p;
	f[u][p] = 1;
	for (int i = 0; i < p; i++) f[u][i] = 0;
	chkmax(ans, p);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int u, v; scanf("%d%d", &u, &v);
		gra[u].push_back(v), gra[v].push_back(u);
	}
	dfs(1, 0);
	printf("%d\n", ans);
	return 0;
}