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>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define FORR(i, n) for(ll i = (ll)n - 1LL; i >= 0LL; i--)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) begin(x), end(x)
using namespace std;
using ll = long long;
template <typename T> using V = vector<T>;
constexpr ll Mod = 998244353;
constexpr ll mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;
const double PI = acos(-1);
template <typename T> constexpr bool chmax(T &a, const T &b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T &b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

int n;
vector<int> G[100009];
int dp[100009];
int m;

bool solve(int v, int p) {
    vector<int> a;
    if((int)G[v].size() & 1) a.push_back(0);
    for(int u : G[v]) {
        if(u == p) continue;
        if(!solve(u, v)) return false;
        if(dp[u] != -1) a.push_back(dp[u] + 1);
    }
    sort(ALL(a));
    int sz = (int)a.size();
    if(sz & 1) {
        int ng = -1, ok = sz;
        while(ok - ng > 1) {
            int x = (ng + ok) >> 1;
            int i = 0, j = sz - 1;
            bool b = true;
            while(i < j) {
                if(i == x) i++;
                if(j == x) j--;
                if(a[i] + a[j] > m) {
                    b = false;
                    break;
                }
                i++;
                j--;
            }
            (b ? ok : ng) = x;
        }
        if(ok == sz) return false;
        dp[v] = a[ok];
        return true;
    } else {
        rep(i, sz) if(a[i] + a[sz - 1 - i] > m) return false;
        dp[v] = -1;
        return true;
    }
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> n;
    rep(i, n - 1) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    int ans = 0;
    rep(i, n) ans += (int)G[i].size() & 1;
    cout << ans / 2 << " ";
    int ng = 0, ok = n;
    while(ok - ng > 1) {
        m = (ng + ok) >> 1;
        (solve(0, -1) ? ok : ng) = m;
    }

    cout << ok << endl;

    return 0;
}