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;

const int maxn = 100100;

int n,A,B,dp[maxn];
vector<int> g[maxn];

bool visit(int u,int pa) {
    vector<int> temp;
  //  cout<<u<<endl;
    if (g[u].size()%2) temp.push_back(-1);
    for (int j=0;j<g[u].size();j++) {
        int v = g[u][j];
        if (v!=pa) {
            if (!visit(v,u)) return false;
            if (dp[v]!=-1) temp.push_back(dp[v]);
        }
    }
    sort(temp.begin(),temp.end());
   // cout<<u<<" "<<temp.size()<<endl;
   // if (temp.size()>0) for (int i=0;i<temp.size();i++) cout<<temp[i]<<" ";
   // cout<<endl;
  //  cout<<"-------------"<<endl;
    if (temp.size()%2) {
        int ll=-1,rr=temp.size();
        while (ll+1<rr) {
            int x = (ll+rr)/2;
            bool ok = true;
            int p1=0,p2=temp.size()-1;
           // cout<<x<<endl;
            while (p1<p2) {
                if (p1==x) p1++;
                if (p2==x) p2--;
                if (temp[p1] + temp[p2] + 2 > B) {
                    ok = false;
                    break;
                }
                p1++;
                p2--;
            }
            if (ok) {
                rr = x;
            }
            else ll = x;
        }
       // cout<<rr<<endl;
        if (rr==temp.size()) return false;
        dp[u] = temp[rr] + 1;
        return true;
    }
    else {
        for (int i=0;i<temp.size();i++)
            if (temp[i] + temp[temp.size()-i-1] + 2 > B) return false;
        dp[u]=-1;
        return true;
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
   // freopen("inp.txt","r",stdin);
    cin>>n;
    for (int i=1;i<n;i++) {
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    A=0;
    for (int i=1;i<=n;i++) A+=g[i].size()%2;
    cout<<A/2<<" ";
    int l=-1,r=n+1;
    while (l+1<r) {
        B = (l+r)/2;
        if (visit(1,-1)) r = B;
        else l = B;
    }
    cout<<r;
}