#include<bits/stdc++.h>

using namespace std;

#define N 120000

int n,ans,f[N];
vector<int> g[N];

void dfs(int u,int fa){
	int mx=0;
	for (int v:g[u])
		if (v!=fa){
			dfs(v,u);
			for (int i=0;i<20;++i)
				if ((f[u]&f[v])>>i&1) mx=max(mx,i+1);
			f[u]|=f[v];
		}
	for (;f[u]>>mx&1;++mx);
	f[u]=(f[u]>>mx|1)<<mx;
	ans=max(ans,mx);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<n;++i){
		int x,y; scanf("%d%d",&x,&y);
		g[x].push_back(y); g[y].push_back(x);
	}
	dfs(1,0);
	printf("%d\n",ans);
	
	return 0;
}
			
### Prompt

Your challenge is to write a CPP solution to the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>

using namespace std;

#define N 120000

int n,ans,f[N];
vector<int> g[N];

void dfs(int u,int fa){
	int mx=0;
	for (int v:g[u])
		if (v!=fa){
			dfs(v,u);
			for (int i=0;i<20;++i)
				if ((f[u]&f[v])>>i&1) mx=max(mx,i+1);
			f[u]|=f[v];
		}
	for (;f[u]>>mx&1;++mx);
	f[u]=(f[u]>>mx|1)<<mx;
	ans=max(ans,mx);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<n;++i){
		int x,y; scanf("%d%d",&x,&y);
		g[x].push_back(y); g[y].push_back(x);
	}
	dfs(1,0);
	printf("%d\n",ans);
	
	return 0;
}
			
```