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;

const int MN = 101234;
int mx = 0;
int dp[MN], col[MN];
vector <int> adj[MN];

int LogDo(int x) {
	int ans = -1;
	for (int i = 0; i < 30; i++)
		if (x >> i & 1)
			ans = i;
	return ans;
}

void dfs(int v = 0, int p = 0) {
	int ans = 0, pre = 0;
	for (int u: adj[v]) {
		if (u == p) 
			continue;
		dfs(u, v);
		int msk = (dp[u] >> col[u] | 1) << col[u];
		ans = max(ans, LogDo(msk & dp[v] & -1) + 1);
		dp[v] |= msk;
		while (dp[v] >> ans & 1)
			ans++;
	}
	col[v] = ans;
	mx = max(mx, ans);
}

int main() {
	int n;
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		adj[v].push_back(u);
		adj[u].push_back(v);
	}
	dfs();
//	for (int i = 0; i < n; i++)
//		cerr << dp[i] << '&' << col[i] << endl;
	cout << mx;
}
