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 mp make_pair
#define pb push_back
#define fst first
#define snd second
#define int long long
#define fast_io() ios::sync_with_stdio(false)
#define FOR(i, l, r) for(int i = (l); i < (r); i++)

typedef long long ll;
typedef pair<ll, ll> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef set<int> si;
typedef double ld;
typedef pair<ld, ld> dd;

const ll INF = 1000000000000000000LL;
const int NMAX = 1e5+4;
const int mod = 1e9+7;
const ld eps = 1e-10;
const ld PI = acos(-1);

int N, A, B;
vi adj[NMAX];


int findA(int u, int p){
    int res = adj[u].size()/2;
    if(u != 0 and adj[u].size() % 2 == 0) res --;
    for(int v : adj[u]) if(v != p) res += findA(v, u);
    return res;
}

int check(vi &v, int excl){
    for(int i = 0, j = v.size()-1; i < j; i++, j--){
        if (i == excl) i++;
        if (j == excl) j--;
        if(i >= j) continue;
        if(v[i] + v[j] > B) return false;
    }
    return true;
}

int leastUp(int u, int p){
    vi vals;
    for(int v : adj[u]) if (v != p) vals.pb(leastUp(v, u) + 1);
    //cout << u << ": ";
    //for(int k : vals) cout << k << " ";
    //cout << endl;
    if(vals.size() == 0) return 0;
    if(vals.size()%2 == 0) {
        if( u == 0 ){
            sort(vals.begin(), vals.end());
            for(int i = 0, j = vals.size()-1; i < j; i++, j--) if(vals[i] + vals[j] > B) return B+1;
            return 0;
        }
        vals.pb(0);
    }
    sort(vals.begin(), vals.end());
    if(vals.back() > B) return B+1;
    int lb = -1, rb = vals.size();
    while(lb+1 != rb) {
        int mb = (lb+rb)/2;
        if(check(vals, mb)) rb = mb;
        else lb = mb;
    }
    if(rb == vals.size()) return B+1;
    //cout << "leastUp " << u << ": " << vals[rb] << endl;
    return vals[rb];
}

int ok(int b){
    B = b;
    //cout << "B = " << b << endl;
    int l = leastUp(0, 0);
    //cout << "l = " << l << " B = " << B << endl;
    return l <= B;
}

signed main(){
    fast_io();
    cin >> N;
    adj[0].pb(0);
    FOR(i, 0, N-1){
        int a, b;
        cin >> a >> b;
        a--; b--;
        adj[a].pb(b);
        adj[b].pb(a);
    }
    A = findA(0, 0);
    int lb = 0, rb = N;
    while(lb + 1 != rb) {
        int mb = (lb+rb)/2;
        if(ok(mb)) rb = mb;
        else lb = mb;
    }

    cout << A << " " << rb << endl; 
    
}