Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

7
1 2
2 3
2 4
4 6
6 7
7 5

Sample Output 1

2

A tree of uninity 1 consisting of vertices 1, 2, 3 and 4 can be constructed from the following: a tree of uninity 0 consisting of vertex 1, a tree of uninity 0 consisting of vertex 3, a tree of uninity 0 consisting of vertex 4, and vertex 2.

A tree of uninity 1 consisting of vertices 5 and 7 can be constructed from the following: a tree of uninity 1 consisting of vertex 5, and vertex 7.

A tree of uninity 2 consisting of vertices 1, 2, 3, 4, 5, 6 and 7 can be constructed from the following: a tree of uninity 1 consisting of vertex 1, 2, 3 and 4, a tree of uninity 1 consisting of vertex 5 and 7, and vertex 6.

Sample Input 2

12
1 2
2 3
2 4
4 5
5 6
6 7
7 8
5 9
9 10
10 11
11 12

Sample Output 2

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16000000")
using namespace std;

typedef pair <int, int> ii;

const int Maxn = 100005;

int n;
vector <int> neigh[Maxn];
int par[Maxn];
bool er[Maxn];

ii Get(int v, int p)
{
    par[v] = p;
    ii res = ii(1, v);
    for (int i = 0; i < neigh[v].size(); i++) {
        int u = neigh[v][i];
        if (u == p || er[u]) continue;
        ii g = Get(u, v);
        res = max(res, ii(g.first + 1, g.second));
    }
    return res;
}

int Solve(int v)
{
    ii g = Get(v, 0);
    g = Get(g.second, 0);
    int tk = g.first / 2;
    int u = g.second;
    while (tk--) u = par[u];
    er[u] = true;
    int mx = 0;
    for (int i = 0; i < neigh[u].size(); i++) {
        int nxt = neigh[u][i];
        if (!er[nxt]) mx = max(mx, Solve(nxt) + 1);
    }
    return mx;
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        int a, b; scanf("%d %d", &a, &b);
        neigh[a].push_back(b);
        neigh[b].push_back(a);
    }
    printf("%d\n", Solve(1));
    return 0;
}