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;

int N;
vector<vector<int> > edges;

vector<int> dp;

int dfs(int now, int from, int len)
{
    multiset<int> l;

    for (auto v : edges[now]) if (v != from)
    {
        if (dfs(v, now, len)) return 1;
        l.insert(dp[v]);
    }
    if (!(l.size() % 2)) l.insert(0);

    vector<int> rem;
    while (!l.empty())
    {
        int t = *(--l.end());
        l.erase(--l.end());
        if (l.empty())
        {
            rem.push_back(t);
            break;
        }
        auto itr = l.upper_bound(len - t);
        if (itr == l.begin()) rem.push_back(t);
        else l.erase(--itr);
    }
    if (rem.size() > 1) return 1;

    dp[now] = rem[0] + 1;
    return 0;
}

bool isAble(int b)
{
    dp = vector<int>(N, -1);
    if (dfs(0, -1, b) || dp[0] > b+1 || (edges[0].size() % 2 == 0 && dp[0] > 1)) return false;
    return true;
}

int main()
{
    cin >> N;
    edges.resize(N);
    for (int i=0; i<N-1; i++)
    {
        int a, b;
        cin >> a >> b;
        a--; b--;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }
    int A = 0;
    for (int i=0; i<N; i++) if ((int)edges[i].size() & 1) A++;
    A /= 2;

    cout << A << " ";

    int l = 0, r = N-1;
    while (r - l >1)
    {
        int c = (l + r) / 2;
        if (isAble(c)) r = c;
        else l = c;
    }
    cout << r << endl;
}
