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, X, a;
vector<vector<int> > adj;

void dfs0(int u, int p) {
    int cnt = 0;
    for(int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i];
        if(v == p) continue;
        cnt++;
        dfs0(v, u);
    }
    a -= cnt / 2;
    if(p != -1 && cnt % 2 == 0) a++;
}

int dfs(int u, int p) {
    multiset<int> st;
    vector<int> tmp;
    for(int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i];
        if(v == p) continue;
        int t = dfs(v, u);
        if(t == -1) return -1;
        if(t + 1 > X) return -1;
        st.insert(t + 1);
        tmp.push_back(t + 1);
    }

    sort(tmp.begin(), tmp.end());

    if((p == -1) ^ (st.size() % 2)) {
        if(p != -1) {
            for(int i = 0; i < (int)tmp.size() - 1; i++) {
                if(tmp[i] + tmp[ (int)tmp.size() - 2 - i ] > X) return -1;
            }
        }
        else {
            for(int i = 0; i < (int)tmp.size(); i++) {
                if(tmp[i] + tmp[ (int)tmp.size() - 1 - i ] > X) return -1;
            }
        }
        while(st.size() >= 2) {
            multiset<int>::iterator it1 = st.end();
            it1--;
            int v = *it1;
            st.erase(it1);

            multiset<int>::iterator it2 = st.upper_bound(X - v);
            if(it2 == st.begin()) return v;
            it2--;
            st.erase(it2);
        }
        return st.size()? *st.begin() : 0;
    }
    else {
        if(p != -1) {
            for(int i = 0; i < (int)tmp.size() - 2; i++) {
                if(tmp[i] + tmp[ (int)tmp.size() - 3 - i ] > X) return -1;
            }
        }
        else {
            for(int i = 0; i < (int)tmp.size() - 1; i++) {
                if(tmp[i] + tmp[ (int)tmp.size() - 2 - i ] > X) return -1;
            }
        }
        int chk = 0;
        while(st.size() >= 2) {
            multiset<int>::iterator it1 = st.end();
            it1--;
            int v = *it1;
            st.erase(it1);

            multiset<int>::iterator it2 = st.upper_bound(X - v);
            if(it2 == st.begin()) {
                if(chk) return v;
                chk = 1;
                continue;
            }
            it2--;
            st.erase(it2);
        }

        return st.size()? *st.begin() : 0;
    }
}

bool f(int x) {
    X = x;
    return dfs(0, -1) != -1;
}

int main() {
    scanf("%d", &N);

    adj.resize(N);
    for(int i = 0; i < N - 1; i++) {
        int u, v; scanf("%d %d", &u, &v);
        u--; v--;

        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    a = 0;
    dfs0(0, -1);

    int s = 1, e = N, b;
    while(s <= e) {
        int m = (s + e)>>1;

        if(f(m)) {
            b = m;
            e = m - 1;
        }
        else s = m + 1;
    }
    printf("%d %d", a, b);
}
