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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);
//const ll mod = 1000000007;
int N;
vector<int> paths[100500];
int A, B;

int dfs(int border, int now, int from) {
    multiset<int> st;
    for(auto to : paths[now]) {
        if(to == from) continue;
        int tmp = dfs(border, to, now);
        if(tmp == -1) return -1;
        st.insert(tmp + 1);
    }
    if(paths[now].size() % 2 == 1) st.insert(0);
    int ansval = -1;
    /*
    cerr << "--" << now << "--" << endl;
    for(auto val: st) cerr << val << " ";
    cerr << endl;
    */
    auto itr = st.end();
    while(true) {
        if(itr == st.begin()) break;
        itr--;
        int val = *itr;
        itr = st.erase(itr);
        auto itr2 = st.upper_bound(border - val);
        if(itr2 == st.begin()) {
            if(ansval != -1) {
                //cerr << "NG: " << now << endl;;
                return -1;
            }
            else {
                ansval = val;
                continue;
            }
        }
        itr2--;
        if(itr == itr2) {
            if(itr == st.end()) itr--;
            else itr++;
        }
        itr2 = st.erase(itr2);
    }
    //cerr << ansval << endl;
    if(now == 0) return 0;
    else return ansval;
}

bool f(int border) {
    //cerr << "-----" << border << "--------" << endl;
    int ret = dfs(border, 0, -1);
    if(ret == 0) return true;
    else return false;
}


int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    for(int i = 1; i <= N - 1; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        paths[a].push_back(b);
        paths[b].push_back(a);
    }
    A = 1;
    for(int i = 0; i < N; i++) {
        A += (paths[i].size() - 1) / 2;
    }
    int ok = 1e6;
    int ng = 0;
    while(ok - ng > 1) {
        int mid = (ok + ng) / 2;
        if(f(mid)) ok = mid;
        else ng = mid;
    }
    B = ok;
    cout << A << " " << B << endl;
    return 0;
}