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 <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <functional>
#define fs first
#define se second

using namespace std;
typedef long long llong;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<llong, llong> pll;
typedef pair<ld, ld> pdd;

int n, m;
vector<int> edge[100001];

int dfs(int x, int p) {
    vector<int> child;
    for (int i : edge[x]) {
        if (i == p) continue;
        int ret = dfs(i, x);
        if (ret > m) return m + 5;
        child.push_back(ret);
    }
    if ((child.size() & 1) == 0) child.push_back(0);
    sort(child.begin(), child.end());
    int s = 0, e = child.size();
    while (s < e) {
        int md = (s + e) / 2, pos = 1;
        for (int i = 0, j = child.size() - 1; ; ++i, --j) {
            if (i == md) ++i;
            if (j == md) --j;
            if (j <= i) break;
            if (child[i] + child[j] > m) {
                pos = 0;
                break;
            }
        }
        if (pos) e = md;
        else s = md + 1;
    }
    if (s == child.size()) return m + 5;
    if (x == 1 && (edge[x].size() & 1) == 0 && s > 0) return m + 5;
    return child[s] + 1;
}

int check() {
    return dfs(1, 0) <= m + 1;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for (int i = 1; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        edge[x].push_back(y);
        edge[y].push_back(x);
    }
    int A = n - 1;
    for (int i = 1; i <= n; ++i) A -= edge[i].size() >> 1;
    int s = 1, e = n - 1;
    while (s < e) {
        m = (s + e) / 2;
        if (check()) e = m;
        else s = m + 1;
    }
    printf("%d %d\n", A, s);
    return 0;
}