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;
 
typedef long long LL;
typedef pair<int, int> II;
 
const int MAXN = (int) 1e5 + 10;
int n;
vector<int> adj[MAXN];
 
int DFS(int u, int T, int par = -1) {
    vector<int> a;
    for (int i = 0; i < (int) adj[u].size(); ++i) {
        int v = adj[u][i];
        if (v != par) a.push_back(DFS(v, T, u));
    }
    if ((int) a.size() % 2 == 0 && par != -1) a.push_back(0);
    sort(a.begin(), a.end());
 
    if ((int) a.size() % 2 == 0) {
        for (int i = 0; i < (int) a.size(); ++i)
            if (a[i] + a[(int) a.size() - i - 1] > T) return T + 3;
        return 0;
    }
 
    int l = 0, r = (int) a.size() - 1, f = -1;
    while (l <= r) {
        int mid = (l + r) >> 1;
        vector<int> b;
        for (int i = 0; i < (int) a.size(); ++i)
            if (i != mid) b.push_back(a[i]);
        bool check = true;
        for (int i = 0; i < (int) b.size() / 2; ++i) {
            if (b[i] + b[(int) b.size() - i - 1] > T) {
                check = false;
                break;
            }
        }
        if (check) {
            f = mid;
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    if (f == -1) {
        // printf("#%d: %d\n", u, T + 3);
        return T + 3;
    }
    // printf("#%d: %d\n", u, a[f] + 1);
    return a[f] + 1;
}
 
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n - 1; ++i) {
        int u, v; scanf("%d%d", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
 
    int l = 1, r = n, f = -1;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (DFS(1, mid) <= mid + 1) {
            f = mid;
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
 
    int cnt = 0;
    for (int i = 1; i <= n; ++i) cnt += (int) adj[i].size() % 2;
 
    printf("%d %d\n", cnt / 2, f);
    return 0;
}