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 N(222222);
int vst[N], msk[N], label[N];
vector<int> adj[N];
void dfs(int v) {
	vst[v] = 1;
	msk[v] = 0;
	int ban(0);
	for(int y : adj[v]) {
		if(vst[y]) continue;
		dfs(y);
		int k(msk[y] & (~((1 << label[y]) - 1)));
		for(int j(0); j <= 30; j++)
			if(msk[v] >> j & 1)
				if(k >> j & 1) 
					ban |= 1 << j;
		msk[v] |= k;
	}
	while((msk[v] >> label[v] & 1) || (1 << label[v]) <= ban) {
		label[v]++;
	}
	msk[v] |= 1 << label[v];
}
int main() {
	int n;
	scanf("%d", &n);
	for(int i(0); i < n - 1; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		adj[x].push_back(y);
		adj[y].push_back(x);
	}
	dfs(1);
	int ans(0);
	while(msk[1]) {
		ans++;
		msk[1] /= 2;
	}
	cout << ans-1 << endl;
}
