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>
#define fi first
#define se second
#define sz(x) ((int)(x).size())
using namespace std;
using pii = pair<int,int>;

int n,L;
pii dp[100001];
vector<int> adj[100001];

void dfs(int cur,int p) {
    dp[cur] = pii(0,0);
    multiset<int> tr;
    for(auto &it:adj[cur]) if(it!=p) {
        dfs(it,cur);
        dp[cur].fi += dp[it].fi;
        if(dp[it].se+1==L) dp[cur].fi++;
        else tr.insert(dp[it].se+1);
    }

    int mx=0;
    while(sz(tr)>1) {
        auto it=prev(tr.end());
        auto jt=tr.upper_bound(L-*it);
        if(jt==tr.begin()) {
            if(mx) dp[cur].fi++;
            mx = *it;
        } else {
            jt--;
            if(jt==it) jt--;
            dp[cur].fi++;
            tr.erase(jt);
        }
        tr.erase(it);
    }
    if(!tr.empty()) {
        if(mx) dp[cur].fi++;
        mx = *tr.begin();
    }
    dp[cur].se = mx;
}

int main() {
    scanf("%d",&n);
    for(int i=1,u,v;i<n;i++) {
        scanf("%d%d",&u,&v);
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    if(n==2) {
        puts("1 1");
        return 0;
    }

    L=n;

    int root=0;
    for(int i=1;i<=n;i++) if(sz(adj[i])>1) {
        root=i;
        break;
    }
    dfs(root,0);
    
    int dt=dp[root].first+(dp[root].se>0);
    int lo=0,up=n;
    while(up-lo>1) {
        L=(lo+up)>>1;
        dfs(root,0);
        if(dp[root].fi+(dp[root].se>0) > dt) lo=L;
        else up=L;
    }
    printf("%d %d\n",dt,up);

    return 0;
}