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 <bits/stdc++.h>
#define MOD 1000000007
using namespace std;
vector<int> V[100005];
int dp[100005];
int ans;int ok;
void dfs(int n,int mid,int p=-1){
    set<pair<int,int> > S;
    for(int i=0;i<V[n].size();++i){
        if(V[n][i]==p) continue;
        dfs(V[n][i],mid,n);
        S.insert(make_pair(dp[V[n][i]],V[n][i]));
    }
    if(S.empty()){
        dp[n]=1;return;
    }

  //  cout << n << " " << S.size() << " size" <<   endl;
    auto it=S.begin();
    bool flag=0;
    while(S.size()>1){
        it=S.end();--it;
       // cout << (*it).first << " " << (*it).second << " pair" << endl;
        auto ip=S.upper_bound(make_pair(mid-1-(*it).first,1000000000));
        if(ip==S.begin()){
            //cout << "A" << endl;
            S.erase(it);--ok;continue;
        }
        --ip;
        if(ip==it){
            if(ip==S.begin()){
               // cout << "B" << endl;
               --ok;S.erase(ip);
               continue;
            }
            --ip;

        }

        S.erase(it);S.erase(ip);--ok;

    }


   dp[n]=(*S.begin()).first+1;
   if(dp[n]>mid){
    dp[n]=2;--ok;
   }
 //  cout << n << " " << dp[n] << endl;




}
void dfs2(int n,int p=-1){
    int sz=0;
    for(int i=0;i<V[n].size();++i){
        if(V[n][i]==p) continue;
        dfs2(V[n][i],n);
        ++sz;
    }
    ans+=sz/2;

}
int main(){
    int n;cin >> n;
    for(int i=1;i<n;++i){
        int a,b;
        scanf("%d %d",&a,&b);
        V[a].push_back(b);V[b].push_back(a);
    }
    int low=1;
    int high=n;
    dfs2(1);
    if(V[1].size()%2) ++ans;
    while(high-low>1){

       ok=ans;
        int mid=(high+low)/2;
       // cout << mid << endl;
        dfs(1,mid);
        if(ok>0){
            high=mid;
        }
        else{
            low=mid+1;
        }
    }
    ok=ans;
    dfs(1,low);
    if(ok>0){
        cout << ans << " " << low-1 << endl;
    }
    else
        cout << ans << " " << high-1 << endl;


    return 0;
}