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 maxn = 100000;
int n, f[maxn + 10];
vector<int> g[maxn + 10];

void dfs(int p, int fa) {
	int cnt[20]; memset(cnt, 0, sizeof cnt);
	for (int i = 0; i < g[p].size(); ++i) {
		int e = g[p][i];
		if (e != fa) {
			dfs(e, p);
			for (int j = 0; j < 20; ++j) cnt[j] += (f[e] >> j & 1);
		}
	}
	int now;
	for (int i = 19; i >= 0; --i)
		if (cnt[i] > 1) break; else if (!cnt[i]) now = i;
	for (int i = 0; i <= 19; ++i)
		if (i == now) f[p] |= 1 << i;
		else if (i > now) f[p] |= (bool(cnt[i]) * (1 << i));
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) {
		int l, r; scanf("%d%d", &l, &r);
		g[l].push_back(r); g[r].push_back(l);
	}
	dfs(1, 0);
	for (int i = 19; i >= 0; --i)
		if (f[1] >> i & 1) {
			printf("%d", i); return 0;
		}
}
