Problem: 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.

Examples

Input

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


Output

3 2


Input

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


Output

2 5


Input

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


Output

3 4

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

using Graph=vector<vector<int>>;

int dfsA(int v,int pre,Graph &g){
    int sum=0;
    int res=0;
    for(int i=0;i<g[v].size();i++){
        if(g[v][i]!=pre){
            sum++;
            res+=dfsA(g[v][i],v,g);
        }
    }
    if(v==0){
        return (sum+1)/2+res;
    }
    else{
        return sum/2+res;
    }
}

int dfsB(int v,int pre,int bound,Graph &g){
    vector<int> len;
    for(int i=0;i<g[v].size();i++){
        if(g[v][i]!=pre) len.push_back(dfsB(g[v][i],v,bound,g)+1);
    }
    if(len.size()==0) return 0;
    sort(len.begin(),len.end());
    if(v!=0){
        if(len.size()%2==0){
            bool isbad=false;
            for(int i=0;i<len.size();i++){
                isbad|=(len[i]+len[len.size()-1-i]>bound);
            }
            if(!isbad) return 0;
            if(len.back()>bound) return bound+1;
            len.pop_back();
        }
        int lb=-1;
        int ub=len.size();
        while(ub-lb>1){
            int mid=(ub+lb)/2;
            int l=-1,r=len.size();
            bool isbad=false;
            while(true){
                if(l+1!=mid) l++;
                else l+=2;
                if(r-1!=mid) r--;
                else r-=2;
                assert(l!=r);
                if(len.size()<=l) break;
                isbad|=(len[l]+len[r]>bound);
            }
            if(isbad) lb=mid;
            else ub=mid;
        }
        if(ub==len.size()) return bound+1;
        else return len[ub];
    }
    else{
        if(len.size()%2==1){
            if(len.back()>bound) return bound+1;
            len.pop_back();
        }
        bool isbad=false;
        for(int i=0;i<len.size();i++){
            isbad|=(len[i]+len[len.size()-1-i]>bound);
        }
        if(isbad) return bound+1;
        else return 0;
    }
}
int main(){
    int n;
    cin>>n;
    Graph g(n);
    for(int i=0;i<n-1;i++){
        int a,b;
        cin>>a>>b;
        a--,b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    int A=dfsA(0,-1,g);
    cout<<A<<" ";
    int lb=0;
    int ub=n;
    while(ub-lb>1){
        int mid=(ub+lb)/2;
        int res=dfsB(0,-1,mid,g);
        if(res<=mid){
            ub=mid;
        }
        else{
            lb=mid;
        }
    }
    cout<<ub<<endl;
    return 0;
}