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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <array>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

const int N = 100010;
VI e[N];

bool good;
int c;

int iso(VI &a){
    int n = a.size();
    int ok = n-1, ng = -1;
    while (ok - ng > 1){
        int mi = (ok + ng) / 2;
        bool f = true;
        REP(i,n/2){
            int x = i;
            if (x >= mi) x++;
            int y = n - 1 - i;
            if (y <= mi) y--;
            if (a[x] + a[y] > c) f = false;
        }
        if (f) ok = mi;
        else ng = mi;
    }
    return a[ok];
}

int dfs(int now, int past){
    if (!good) return 0;
    VI a;
    for (int next : e[now]){
        if (next == past) continue;
        a.push_back(dfs(next, now) + 1);
    }
    sort(ALL(a));
    int n = a.size();
    // cout << now << " " << n << endl;
    if (n == 0) return 0;
    if (a.back() > c) good = false;
    if (n % 2 == 0){
        // cout << "even" << endl;
        bool f = true;
        REP(i,n/2){
            if (a[i] + a[n-1-i] > c) f = false;
        }
        if (f) return 0;
        a.pop_back();
        return iso(a);
    }
    REP(i,n/2){
        if (a[i] + a[n-2-i] > c) good = false;
    }
    return iso(a);
}

int main() {
    int n;
    cin >> n;
    REP(i,n-1){
        int a, b;
        scanf("%d %d", &a, &b);
        a--;b--;
        e[a].push_back(b);
        e[b].push_back(a);
    }

    int a = 0;
    REP(i,n) a += (e[i].size() % 2);
    a = (a + 1) / 2;

    int root;
    REP(i,n) if (e[i].size() == 1) root = i;

    // cout << "root" << root << endl;

    int ok = n, ng = 0;
    while (ok - ng > 1){
        c = (ok + ng) / 2;
        // cout << c << endl;
        good = true;
        dfs(root, -1);
        if (good) ok = c;
        else ng = c;
    }


    cout << a << " " << ok << endl;

    return 0;
}