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 = 100005;

struct node {
	int to, next;
} edge[MAXN*2];
int head[MAXN], top = 0;
inline void push(int i, int j)
{ edge[++top] = (node) {j, head[i]}, head[i] = top; }

int n, a, b;

int dp[MAXN], lg[MAXN];

inline int lowbit(int i)
{ return i&(-i); }

int lab[MAXN];

void dfs(int nd, int f)
{
	int past = 0, minv = 0;
	for (int i = head[nd]; i; i = edge[i].next) {
		int to = edge[i].to;
		if (to == f) continue;
		dfs(to, nd);
		minv = max(minv, lg[past&dp[to]]);
		past |= dp[to];
	}
	past >>= minv;
	lab[nd] = lg[lowbit(past+1)]+minv;
	dp[nd] = (past+1)<<minv;
}

int main()
{
	scanf("%d", &n);
	lg[1] = 0;
	for (int i = 2; i <= n; i++) lg[i] = lg[i>>1]+1;
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &a, &b);
		push(a, b), push(b, a);
	}
	dfs(1, 0);
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		ans = max(ans, lab[i]);
	}
	cout << ans << endl;
	return 0;
}
