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 N = 100005;

int n, A, newA;
vector<int> G[N];
int lim;

int findA(int u, int p) {
    int rem = 0;
    for (int v : G[u]) if (v != p) {
        rem += findA(v, u);
    }
    A += rem / 2; rem %= 2;
    if (rem == 1) return rem;
    return (u == 1) ? 0 : 1;
}

int dfs(int u, int p) {
    multiset<int> s;
    for (int v : G[u]) if (v != p) {
        int cur = dfs(v, u) + 1;
        s.insert(cur);
    }

    vector<int> rem;
    while(s.size()) {
        multiset<int>::iterator it = s.end(); --it;
        int x = (*it);
        s.erase(it);
        it = s.upper_bound(lim - x);
        if (it == s.begin()) {
            rem.push_back(x);
        } else --it, s.erase(it), ++newA;
    }

    reverse(rem.begin(), rem.end());
    while(rem.size() && rem.back() == lim) ++newA, rem.pop_back();
    if (rem.empty()) return 0;
    else { newA += rem.size() - 1; return rem[0]; }
}

bool check() {
    newA = 0;
    newA += dfs(1, -1) != 0;
    return newA <= A;
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> n;
    for (int i = 1; i < n; ++i) {
        int u, v; cin >> u >> v;
        G[u].push_back(v); G[v].push_back(u);
    }
    A += findA(1, -1);

    int l = 1, r = n;
    while(l < r) {
        lim = ((l + r) >> 1);
        if (check()) r = lim; else l = lim + 1;
    }
    cout << A << ' ' << l << endl;
}