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;

int N;
vector<int> G[100010];
int ans;

int DFS1(int v,int p)
{
    int cnt = 0,sum = 0;
    for(int i = 0; i < G[v].size(); i++)
    {
        int ne = G[v][i];
        if(ne == p)continue;
        sum += DFS1(ne,v);
        cnt++;
    }
    if(cnt == 0)return 1;
    return sum - (cnt - 1) / 2;
}

int DFS2(int v,int p,int x)
{
    vector<int> V;
    int cnt = 0;
    for(int i = 0; i < G[v].size(); i++)
    {
        int ne = G[v][i];
        if(ne == p)continue;
        V.push_back(DFS2(ne,v,x));
        cnt++;
    }
    if(cnt == 0)
    {
        return 1;
    }
    sort(V.begin(),V.end());
    if(V.back() > x)
    {
        return 0;
    }
    if(V[0] == 0)
    {
        return 0;
    }
    if(cnt % 2 == 0)
    {
        for(int i = 0; i < V.size() / 2 + 1; i++)
        {
            if(i == V.size() / 2)
            {
                return 1;
            }
            if(V[i] + V[V.size() - 1 - i] > x)
            {
                if(V.back() > x)
                {
                    return 0;
                }
                V.pop_back();
                break;
            }
        }
    }
    int l = -1,r = V.size() - 1;
    while(r - l > 1)
    {
        int mid = (l + r) / 2;
        int L = 0,R = V.size() - 1;
        for(int i = 0; i <= V.size() / 2; i++,L++,R--)
        {
            if(i == V.size() / 2)
            {
                r = mid;
                break;
            }
            if(L == mid)L++;
            if(R == mid)R--;
            if(V[L] + V[R] > x)
            {
                l = mid;
                break;
            }
        }
    }

    int L = 0,R = V.size() - 2;
    for(int i = 0; i <= V.size() / 2; i++,L++,R--)
    {
        if(i == V.size() / 2)
        {
            return V[r] + 1;
        }
        if(L == r)L++;
        if(R == r)R--;
        if(V[L] + V[R] > x)
        {
            return 0;
            break;
        }
    }
}

bool judge(int mid)
{
    int res;
    for(int i = 1; i <= N; i++)if(G[i].size() == 1)
    {
        res = DFS2(i,0,mid);
        break;
    }
    return (res != 0 && mid >= res - 1);
}

int main()
{
    scanf("%d",&N);
    for(int i = 0; i < N - 1; i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
    }

    //ans = DFS1(1,0);
    for(int i = 1; i <= N; i++)if(G[i].size() % 2)ans++;
    ans /= 2;

    int l = 0,r = N - 1;
    while(r - l > 1)
    {
        int mid = (l + r) / 2;
        if(judge(mid))r = mid;
        else l = mid;
    }

    printf("%d %d\n",ans,r);
    return 0;
}