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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int INF=1001001001;

int N;
vint G[111111];

int K;

int dfs(int v,int p){
    vint a;
    for(auto u:G[v]){
        if(u==p)continue;
        int tmp=dfs(u,v);
        if(tmp==INF)return INF;
        a.pb(tmp);
    }

    sort(all(a));
    if(p==-1)return 0;
    if(a.size()==0)return 1;

    if(a.size()%2==0){
        bool f=true;
        rep(i,a.size()/2)if(a[i]+a[a.size()-i-1]>K)f=false;
        if(f)return 1;
        a.pop_back();
    }

    int lb=-1,ub=a.size();
    while(ub-lb>1){
        int mid=(ub+lb)/2;

        bool f=true;
        int l=0,r=a.size()-1;
        rep(i,a.size()/2){
            if(l==mid)l++;
            if(r==mid)r--;
            if(a[l]+a[r]>K)f=false;
            l++;r--;
        }
        if(f)ub=mid;
        else lb=mid;
    }
    if(ub==a.size())return INF;
    if(a[ub]+1>K)return INF;
    return a[ub]+1;
}

bool C(int x){
    K=x;
    rep(i,N)if(G[i].size()==1)return dfs(i,-1)!=INF;
}

signed main(){
    cin>>N;
    rep(i,N-1){
        int a,b;
        cin>>a>>b;
        a--;b--;
        G[a].pb(b);G[b].pb(a);
    }

    int cnt=0;
    rep(i,N)if(G[i].size()%2)cnt++;

    int A=cnt/2;

    int lb=0,ub=N;
    while(ub-lb>1){
        int mid=(ub+lb)/2;
        if(C(mid))ub=mid;
        else lb=mid;
    }

    cout<<A<<" "<<ub<<endl;
    return 0;
}