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 <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(x, a, b) for(int x = (a); x <= (b); ++x)
#define rop(x, a, b) for(int x = (a); x < (b); ++x)
#define per(x, a, b) for(int x = (a); x >= (b); --x)
using namespace std;
typedef long long LL;
typedef double DB;
int ft[100005][20], as;
struct node {
	int vs[100005];
	int hd[100005], vr[200005], nt[200005], tt, rt;
	void push(int x, int y) {
		vr[++tt] = y; nt[tt] = hd[x]; hd[x] = tt;
	}
	void dfs(int nw, int en) {
		for(int i = hd[nw]; i; i = nt[i]) {
			if(vr[i] == en) continue;
			dfs(vr[i], nw);
			rop(j, 0, 20) ft[nw][j] += ft[vr[i]][j];
		}
		int x = 0;
		per(j, 19, 0) {
			if(ft[nw][j] > 1) break;
			if(!ft[nw][j]) x = j;
		}
		as = max(as, x);
		rep(j, 0, x) ft[nw][j] = 0;
		ft[nw][x] = 1;
		return;
	}
}T;
int n, x, y;
int main() {
	scanf("%d", &n);
	rop(i, 1, n) {
		scanf("%d%d", &x, &y);
		T.push(x, y);
		T.push(y, x);
	}
	T.dfs(1, 0);
	cout << as << endl; 
	return 0;
}
