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 rep(i, n)  for(long long i=0;i<(long long)(n);i++)
#define REP(i,k,n) for(long long i=k;i<(long long)(n);i++)
#define all(a) a.begin(),a.end()
#define pb emplace_back
#define eb emplace_back
#define lb(v,k) (lower_bound(all(v),k)-v.begin())
#define ub(v,k) (upper_bound(all(v),k)-v.begin())
#define fi first
#define se second
#define pi M_PI
#define PQ(T) priority_queue<T>
#define SPQ(T) priority_queue<T,vector<T>,greater<T>>
#define dame(a) {out(a);return 0;}
#define decimal cout<<fixed<<setprecision(15);
#define dupli(a) a.erase(unique(all(a)),a.end())
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef tuple<ll,ll,ll,ll> PPP;
typedef multiset<ll> S;
using vi=vector<ll>;
using vvi=vector<vi>;
using vvvi=vector<vvi>;
using vp=vector<P>;
using vvp=vector<vp>;
using vb=vector<bool>;
using vvb=vector<vb>;
const ll inf=1001001001001001001;
const int INF=1001001001;
const int mod=1000000007;
const double eps=1e-10;
template<class T> bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<class T> bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
template<class T> void out(T a){cout<<a<<'\n';}
template<class T> void outp(T a){cout<<'('<<a.fi<<','<<a.se<<')'<<'\n';}
template<class T> void outvp(T v){rep(i,v.size())cout<<'('<<v[i].fi<<','<<v[i].se<<')';cout<<'\n';}
template<class T> void outvvp(T v){rep(i,v.size())outvp(v[i]);}
template<class T> void outv(T v){rep(i,v.size()){if(i)cout<<' ';cout<<v[i];}cout<<'\n';}
template<class T> void outvv(T v){rep(i,v.size())outv(v[i]);}
template<class T> bool isin(T x,T l,T r){return (l)<=(x)&&(x)<=(r);}
template<class T> void yesno(T b){if(b)out("yes");else out("no");}
template<class T> void YesNo(T b){if(b)out("Yes");else out("No");}
template<class T> void YESNO(T b){if(b)out("YES");else out("NO");}
template<class T> void noyes(T b){if(b)out("no");else out("yes");}
template<class T> void NoYes(T b){if(b)out("No");else out("Yes");}
template<class T> void NOYES(T b){if(b)out("NO");else out("YES");}
void outs(ll a,ll b){if(a>=inf-100)out(b);else out(a);}
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll modpow(ll a,ll b){a%=mod;if(b==0)return 1;if(b&1)return a*modpow(a,b-1)%mod;ll k=modpow(a,b/2);return k*k%mod;}
ll md;
vvi g;
ll e;
ll dfs(int i,int p){
    ll res=0;
    ll cnt=0;
    for(ll x:g[i])if(x!=p){
        res+=dfs(x,i);
        cnt++;
    }
    if(cnt)res-=(cnt-1)/2;
    else res++;
    return res;
}
ll check(int i,int p){
    vi v;
    for(ll x:g[i])if(x!=p){
        ll t=check(x,i);
        if(t==-1)return -1;
        v.pb(t);
    }
    if(i==e){
        return 1;
    }
    sort(all(v));
    if(!v.size())return 1;
    if(v.size()%2==0){
        bool b=true;
        rep(j,v.size()/2)if(v[j]+v[v.size()-j-1]>md)b=false;
        if(b)return 1;
        v.pop_back();
    }
    if(v[0]>=md)return -1;
    ll t=lb(v,md);
    ll ok=t,ng=-1;
    while(ok-ng>1){
        ll k=(ok+ng)/2;
        bool able=true;
        int a=0,b=v.size()-1;
        rep(j,v.size()/2){
            if(a==k)a++;
            if(b==k)b--;
            if(v[a]+v[b]>md)able=false;
            a++;b--;
        }
        if(able)ok=k;
        else ng=k;
    }
    if(ok==t)return -1;
    return v[ok]+1;
}
int main(){
    ll n;cin>>n;
    g=vvi(n);
    rep(i,n-1){
        ll a,b;cin>>a>>b;a--;b--;
        g[a].pb(b);
        g[b].pb(a);
    }
    int ne=0;
    while(g[ne].size()>1)ne++;
    e=ne;
    ll ans=dfs(ne,-1);
    ll ok=n-1,ng=0;
    while(ok-ng>1){
        md=(ok+ng)/2;
        if(check(ne,-1)!=-1)ok=md;
        else ng=md;
    }
    cout<<ans<<' '<<ok<<endl;
}