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:
/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/

#include <bits/stdc++.h>
using namespace std;

const int mx=1e5+5;
vector <int> edges[mx];
int DP[mx];
bool vis[mx];

bool check (vector <int>& val,int b,int mid)
{
    int i,j;
    bool ret=true;
    i=0;
    j=val.size()-1;
    while (i<j) {
        if (i==mid) i++;
        if (j==mid) j--;
        if (val[i]+val[j]>b) {
            ret=false;
            break;
        }
        else i++,j--;
    }
    return ret;
}

bool DFS (int u,int b)
{
    vis[u]=true;
    bool ret=true;
    vector <int> val;

    for (auto v : edges[u]) {
        if (!vis[v]) {
            ret=(ret & DFS(v,b));
            val.push_back(DP[v]);
        }
    }
    if (!ret) return ret;
    if (val.empty()) {
        DP[u]=1;
        return true;
    }
    int m=val.size();
    if (m%2==0) {
        val.push_back(0);
        m++;
    }
    sort(val.begin(),val.end());

    int hi,lo,mid,ans=-1;
    lo=0,hi=m-1;
    while (lo<=hi) {
        mid=(lo+hi)/2;
        if (val[mid]<=b && check(val,b,mid)) {
            ans=mid;
            hi=mid-1;
        }
        else lo=mid+1;
    }
    if (ans==-1) return false;
    else {
        DP[u]=1+val[ans];
        return true;
    }
}

int main ()
{
    int i,n;

    scanf ("%d",&n);
    for (i=1;i<n;i++) {
        int u,v;
        scanf ("%d %d",&u,&v);
        edges[u].push_back(v);
        edges[v].push_back(u);
    }

    int u,hi,lo,mid,ans,cnt=0;
    for (i=1;i<=n;i++) {
        cnt+=edges[i].size()%2;
        if (edges[i].size()==1) u=i;
    }

    lo=1,hi=n-1;
    while (lo<=hi) {//cout<<lo<<endl;
        mid=(lo+hi)/2;
        memset(vis,0,sizeof(vis));
        if (DFS(u,mid)) {
            ans=mid;
            hi=mid-1;
        }
        else lo=mid+1;
    }
    printf ("%d %d\n",cnt/2,ans);

    return 0;
}