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 <iostream>
#include <vector>
#include <tuple>
#include <algorithm>

#define rep(i,i0,n) for (int (i) = (i0); (i) < (n); ++(i))

using namespace std;

int n = 0;
vector<int> edges[101010];
bool visited[101010];

int dfs(int v, int b, bool r = false) {
    if (!r) {
        fill(visited, visited + 101010, false);
    }

    visited[v] = true;
    vector<int> lengths;

    for (auto c : edges[v]) {
        if (!visited[c]) {
            int k = dfs(c, b, true);
            if (!k || b < k) return 0;
            lengths.push_back(k);
        }
    }

    if (edges[v].size() & 1) {
        lengths.push_back(0);
    }

    int hold = 0;
    sort(lengths.begin(), lengths.end());
    while (1 < lengths.size()) {
        int k = lengths.back();
        lengths.pop_back();
        auto i = upper_bound(lengths.begin(), lengths.end(), b - k);
        if (i == lengths.begin()) {
            if (r && !hold) {
                hold = k;
            } else {
                return 0;
            }
        } else {
            lengths.erase(i - 1);
        }
    }

    if (hold) {
        lengths.push_back(hold);
    }

    if (lengths.size()) {
        return lengths[0] + 1;
    } else {
        return 1;
    }
}


tuple<int, int> solve() {
    int root = -1;
    int a = 1;

    rep(i, 0, n) {
        size_t c = edges[i].size();
        if (3 <= c) {
            a += (c - 1) / 2;
            root = i;
        }
    }

    if (root < 0) {
        return make_tuple(1, n - 1);
    }

    int ok = n - 1;
    int ng = 1;

    while (ng + 1 < ok) {
        int mid = (ok + ng) / 2;
        if (dfs(root, mid)) {
            ok = mid;
        } else {
            ng = mid;
        }
    }

    return make_tuple(a, ok);
}

int main(int argc, const char * argv[]) {
    int a, b;
    
    cin >> n;
    rep(i, 0, n - 1) {
        cin >> a >> b;
        --a;
        --b;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }

    auto p = solve();

    cout << get<0>(p) << ' ' << get<1>(p) << endl;

    return 0;
}