Problem:
Score : 900 points

Problem Statement

Takahashi has decided to make a Christmas Tree for the Christmas party in AtCoder, Inc.

A Christmas Tree is a tree with N vertices numbered 1 through N and N-1 edges, whose i-th edge (1\leq i\leq N-1) connects Vertex a_i and b_i.

He would like to make one as follows:

Specify two non-negative integers A and B.

Prepare A Christmas Paths whose lengths are at most B. Here, a Christmas Path of length X is a graph with X+1 vertices and X edges such that, if we properly number the vertices 1 through X+1, the i-th edge (1\leq i\leq X) will connect Vertex i and i+1.

Repeat the following operation until he has one connected tree:

Select two vertices x and y that belong to different connected components. Combine x and y into one vertex. More precisely, for each edge (p,y) incident to the vertex y, add the edge (p,x). Then, delete the vertex y and all the edges incident to y.

Properly number the vertices in the tree.

Takahashi would like to find the lexicographically smallest pair (A,B) such that he can make a Christmas Tree, that is, find the smallest A, and find the smallest B under the condition that A is minimized.

Solve this problem for him.

Constraints

2 \leq N \leq 10^5

1 \leq a_i,b_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

For the lexicographically smallest (A,B), print A and B with a space in between.

Sample Input 1

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

Sample Output 1

3 2

We can make a Christmas Tree as shown in the figure below:

Sample Input 2

8
1 2
2 3
3 4
4 5
5 6
5 7
5 8

Sample Output 2

2 5

Sample Input 3

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

Sample Output 3

3 4

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<bool, Int> P;


vector<int> edge[110000];
int n, u, v;

P check(vector<int> &tmp, int b){
    int k = tmp.size();
    if(k == 1)return P(true, tmp[0]);
    for(int i = 0;i < k / 2;i++)
        if(tmp[i] + tmp[k-2-i] > b)return P(false, 0);
    for(int i = 0;i < k / 2;i++)
        if(tmp[i] + tmp[k-1-i] > b)return P(true, tmp[k-1-i]);
    for(int i = 0;i < k / 2 + 1;i++)
        if(tmp[i] + tmp[k-i] > b)return P(true, tmp[0]);
    return P(true, tmp[0]);
}

P dfs(int x, int b, int last = -1){
    vector<int> tmp;
    for(auto to: edge[x]){
        if(to == last)continue;
        auto p = dfs(to, b, x);
        if(!p.first)return P(false, 0);
        if(p.second == b)return P(false, 0);
        tmp.push_back(p.second + 1);
    }
    sort(tmp.begin(), tmp.end());
    int k = tmp.size();
    if(tmp.size() % 2 == 0){
        bool ok = true;
        for(int i = 0;i < k/2;i++)
            if(tmp[i] + tmp[k-1-i] > b)ok = false;
        if(ok)return P(true, 0);
        tmp.pop_back();
    }
    return check(tmp, b);
}

int main(){
    cin >> n;
    int a = 0;
    for(int i = 0;i < n-1;i++){
        cin >> u >> v;
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    for(int i = 1;i <= n;i++)if(edge[i].size() % 2)a++;
    a/=2;

    int bottom  = 0, top = n;
    while(top - bottom > 1){
        int mid = (top + bottom) / 2;
        if(dfs(1, mid).first)top = mid;
        else bottom = mid;
    }
    cout << a << " " << top << endl;
    
    return 0;
}