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 <bits/stdc++.h>
using namespace std;

int N;
vector<int> edges[100000];

bool check2(const vector<int>& v, int omit, int K){
    int sz = v.size();
    int i = 0, j = sz-1;
    while(true){
        if(i == omit) i++;
        if(j == omit) j--;
        if(i >= j) return true;
        if(v[i] + v[j] > K) return false;
        i++; j--;
    }
}

int find_opt(const vector<int>& v, int K){
    int sz = v.size();
    int ng = -1, ok = sz;
    while(ok-ng>1){
        int mid = (ng+ok)/2;
        (check2(v, mid, K) ? ok : ng) = mid;
    }
    return (ok == sz ? -1 : v[ok]);
}

int dfs(int i, int par, int K){
    vector<int> v;
    for(auto j : edges[i]){
        if(j == par) continue;
        int ret = dfs(j, i, K);
        if(ret == -1 || ret+1 > K) return -1;
        v.push_back(ret+1);
    }
    if(v.size() % 2 == 0) v.push_back(0);
    sort(v.begin(), v.end());

    return find_opt(v, K);
}

bool check(int K){
    int result = dfs(0, -1, K);
    if(edges[0].size() % 2 == 1){
        return (result != -1);
    }else{
        return (result == 0);
    }
}

int main(){
    cin >> N;
    for(int i=0; i<N-1; i++){
        int a, b;
        cin >> a >> b;
        edges[a-1].push_back(b-1);
        edges[b-1].push_back(a-1);
    }

    int A = 0;
    for(int i=0; i<N; i++){
        if(edges[i].size() % 2 == 1) A++;
    }
    A /= 2;

    int ng = 0, ok = N;
    while(ok-ng>1){
        int mid = (ok+ng)/2;
        (check(mid) ? ok : ng) = mid;
    }
    int B = ok;

    cout << A << " " << B << endl;
    return 0;
}