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 = 100005;
int n;
vector<int> adjlist[N];
int p[N];
int ct[N][20];
int ANS = 0;
void dfs(int u){
    for (auto v: adjlist[u]){
        if (p[u] == v) continue;
        p[v] = u;
        dfs(v);
        for (int i = 0; i < 20; i++){
            ct[u][i] += ct[v][i];
        }
    }
    int ans[20]; memset(ans,0,sizeof(ans));
    int id = 19;
    for (; id >= 0; id--){
        if (ct[u][id] >= 2) break;
    }
    if (id == -1) id = 0;
    for (; id < 20; id++){
        if (ct[u][id] == 0){
            ANS = max(ANS,id);
            ans[id] = 1;
            break;
        }
    }
    for (; id < 20; id++){
        ans[id] += ct[u][id];
    }
    for (int i = 0; i < 20; i++){
        ct[u][i] = ans[i];
    }
    /*printf("%d ",u);
    for (int i = 0; i < 20; i++){
        printf("%d",ct[u][i]);
    }
    printf("\n");*/
}
int main(){
    scanf("%d",&n);
    for (int i = 1,a,b;i < n; i++){
        scanf("%d%d",&a,&b);
        adjlist[a].push_back(b);
        adjlist[b].push_back(a);
    }
    p[1] = 1;
    dfs(1);
    printf("%d ",ANS);
}
