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 rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const double EPS = 1e-10;
const int INF = 100000000;
const ll MOD = 1000000007;

int n;
vector<int> e[100000];
int ansa = 1;

bool checker(vector<int> vec, int mid, int x) {
    int i = 0, j = vec.size()-1;
    bool ok = true;
    rep(_,vec.size()/2) {
        if (i == mid) i++;
        if (j == mid) j--;
        if (vec[i]+vec[j] > x) {
            ok = false;
            break;
        }
        i++; j--;
    }
    return ok;
}

int dfs(int v, int par, int x) {
    vector<int> vec;
    rep(i,e[v].size()) {
        if (e[v][i] == par) continue;
        int z = dfs(e[v][i], v, x);
        if (z == INF) {
            return INF;
        }
        vec.push_back(z);
    }
    if (vec.size() == 0) {
        return 1;
    }
    if (vec.size() == 1) {
        return vec[0]+1;
    }
    sort(vec.begin(), vec.end());
    if (vec.size()%2 == 0) {
        bool ok = true;
        rep(i,vec.size()/2) {
            if (vec[i]+vec[vec.size()-1-i] > x) {
                ok = false;
                break;
            }
        }
        if (ok) {
            return 1;
        }
        if (v == 0) return INF;
        if (vec[vec.size()-1] > x) return INF;
        vec.pop_back();
    }
    int st = -1, en = vec.size(), mid;
    while (en-st > 1) {
        mid = (st+en)/2;
        if (checker(vec, mid, x)) {
            en = mid;
        } else {
            st = mid;
        }
    }
    if (en == vec.size()) {
        return INF;
    }
    return vec[en]+1;
}

bool solve(int x) {
    int z = dfs(0, 0, x);
    if (z-1 > x) return false;
    return true;
}

int main(){
    cin >> n;
    rep(i,n-1) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        e[a].push_back(b);
        e[b].push_back(a);
    }
    if (n == 2) {
        cout << 1 << " " << 1 << endl;
        return 0;
    }
    rep(i,n) {
        if (e[i].size() > 2) {
            ansa += (e[i].size()-2+1)/2;
        }
    }
    int st = 1, en = n, mid;
    while (en-st > 1) {
        mid = (st+en)/2;
        if (solve(mid)) {
            en = mid;
        } else {
            st = mid;
        }
    }
    cout << ansa << " " << en << endl;
}