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 = 1e5 + 10, lg = 19;
int mask[N], cnt[N][lg + 2], ans, n;
vector<int> gr[N];
void dfs(int v, int par = -1) {
	int rn = 0;
	for (auto u : gr[v]) {
		if (u != par) {
			dfs(u, v), mask[v] |= mask[u];
			for (int i = 0; i < lg; i++) {
				if ((mask[u] >> i) & 1)
					cnt[v][i]++;
				if (cnt[v][i] > 1)
					rn = max(rn, i + 1);
			}
		}
	}
	for (int i = rn; i < lg; i++) {
		if (!((mask[v] >> i) & 1)) {
			rn = i;
			break;
		}
	}
	ans = max(ans, rn);
	mask[v] = mask[v] & (((1 << lg) - 1) ^ ((1 << rn) - 1));
	mask[v] |= (1 << rn);
}	
int main() {
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int x, y;
		cin >> x >> y;
		x--, y--;
		gr[x].push_back(y), gr[y].push_back(x);
	}
	dfs(0);
	cout << ans;
	return 0;
}
