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 <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
static_assert(sizeof(long) >= 8, "long too short");

vector<int> adj[100000];
vector<int> data1[100000];

int solvea(int n) {
    int ans = 0;
    for (int i = 0; i < n; i++)
        ans += adj[i].size() % 2;
    return ans / 2;
}

int dfs(int u, int p, int b) {
    auto &v = data1[u];
    v.clear();
    v.reserve(adj[u].size() + 1);

    for (int c : adj[u]) {
        if (c != p) {
            v.push_back(dfs(c, u, b));
        }
    }
    if (adj[u].size() % 2)
        v.push_back(0);
    sort(v.begin(), v.end());

    if (p < 0) {
        int ans = 0;
        for (int i = 0, j = v.size() - 1; i < j; i++, j--)
            ans = max(ans, v[i] + v[j]);
        return ans;
    } else {
        int l = 0, r = v.size();
        while (l < r) {
            int k = (l + r) / 2;
            bool ok = true;
            for (int i = 0, j = v.size() - 1; i < j; i++, j--) {
                if (i == k)
                    i++;
                if (j == k)
                    j--;
                if (i >= j)
                    break;
                if (v[i] + v[j] > b) {
                    ok = false;
                    break;
                }
            }
            if (ok)
                r = k;
            else
                l = k + 1;
        }
        if (r < (int) v.size())
            return v[r] + 1;
        return b + 1;
    }
}

bool checkb(int b) {
    return dfs(0, -1, b) <= b;
}

int solveb(int n) {
    int l = 1, r = n - 1;
    while (l < r) {
        int m = (l + r) / 2;
        if (checkb(m))
            r = m;
        else
            l = m + 1;
    }
    return r;
}

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        a--, b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    int a = solvea(n);
    int b = solveb(n);
    printf("%d %d\n", a, b);
}
