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:
#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
using namespace std;

int dfs(const vector<vector<int> >& edges, int curr, int prev, int limit)
{
    multiset<int> ms;
    for(int next : edges[curr]){
        if(next == prev)
            continue;
        int x = dfs(edges, next, curr, limit);
        if(x == -1)
            return -1;
        ms.insert(x);
    }

    vector<int> v;
    while(!ms.empty()){
        int x = *ms.rbegin();
        ms.erase(--ms.end());

        auto it = ms.upper_bound(limit - 1 - x);
        if(it == ms.begin()){
            v.push_back(x);
        }
        else{
            -- it;
            ms.erase(it);
        }
    }

    if(!v.empty() && limit < v[0] + 1)
        return -1;

    if(v.empty())
        return 1;
    else if(v.size() == 1)
        return v[0] + 1;
    else if(v.size() == 2 && prev != -1)
        return v[1] + 1;
    else
        return -1;
}

pair<int, int> solve(const vector<vector<int> >& edges)
{
    int n = edges.size();
    int ans = 1;
    for(int i=0; i<n; ++i)
        ans += (edges[i].size() - 1) / 2;

    int left = 1;
    int right = n;
    while(left < right){
        int mid = (left + right) / 2;
        if(dfs(edges, 0, -1, mid) != -1)
            right = mid;
        else
            left = mid + 1;
    }

    return make_pair(ans, left - 1);
}

int main()
{
    int n;
    cin >> n;
    vector<vector<int> > edges(n);
    for(int i=0; i<n-1; ++i){
        int a, b;
        cin >> a >> b;
        -- a;
        -- b;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }

    pair<int, int> ans = solve(edges);
    cout << ans.first << ' ' << ans.second << endl;

    return 0;
}