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;
#define MAXN	100005
#define MAXLOG	20
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
vector <int> a[MAXN];
int ans[MAXN][MAXLOG];
void work(int pos, int fa) {
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i] != fa) {
			work(a[pos][i], pos);
			for (int j = 0; j < MAXLOG; j++)
				ans[pos][j] += ans[a[pos][i]][j];
		}
	int home = -1;
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (ans[pos][i] >= 2) {
			home = i;
			break;
		}
	for (int i = 0; i <= home; i++)
		ans[pos][i] = 0;
	for (int i = home + 1; i < MAXLOG; i++) {
		ans[pos][i]++;
		if (ans[pos][i] == 1) break;
		else ans[pos][i] = 0;
	}
}
int main() {
	int n; read(n);
	for (int i = 1; i < n; i++) {
		int x, y;
		read(x), read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	work(1, 0);
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (ans[1][i]) {
			printf("%d\n", i);
			return 0;
		}
	return 0;
}