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
#include <bits/stdc++.h>
typedef int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int , int >
#define F first
#define S second
#define endl '\n'
//#define int long long
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#define kill(x) return cout<<x<<'\n', 0;
using namespace std;
const int N=2e5+100;
vector <int> g[N];
ll dp[N];
ll p1=0;
ll z=0;
void dfs(ll v,ll p,ll k){
    multiset <int> s;
    for (auto u : g[v]){
        if (u==p) continue;
        dfs(u,v,k);
        s.insert(dp[u]+1);
        if (dp[u]+1>k){
            p1=1;
            return ;
        }
    }
    if (p1) return ;
   // if (v==1) cout << s.size() << " end " << endl;
   vector <int> y;
    while(s.size()>1){
        auto t=s.end();
        t--;
        ll val1=*t;
        s.erase(t);
        if (*s.begin()+val1>k){

                y.pb(val1);
                continue;
        }
        t=s.upper_bound(k-val1);
        t--;
        s.erase(t);
        z++;
    }
    if (s.size()){
        y.pb(*s.begin());
    }
    if (y.size()>2){
        p1=1;
        return ;
    }
    if (v==1 && y.size()>1) p1=1;
    if (y.size()){
        sort(y.begin(),y.end());
        for (int i=1;i<y.size();i++) z++;
        dp[v]=y[0];
        if (v==1) z++;
    }
    if (dp[1]>k) p1=1;
}
int32_t main(){
    sync;
    ll n;
    cin >> n;
    for (int i=1;i<n;i++){
        ll u,v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    ll l=0,r=n+10;
    while(r-l>1){
        memset(dp,0,sizeof dp);
        p1=0;
        ll mid=(r+l)/2;
        z=0;
        dfs(1,1,mid);
        if (!p1) r=mid;
        else l=mid;
        //cout << z << " f " << mid << endl;
    }memset(dp,0,sizeof dp);
    z=0;
    p1=0;
    dfs(1,1,r);
    cout << z << " " << r << endl;
}
