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;

vector<int> g[100001];
vector<int> tree[100001];
vector<int> l[100001];
int p[100001];
int bye[100001];

int ans1 = 0;
int lim = 0;

void dfs1(int x){
    int cntt = 0;
    for(auto &nxt: g[x]){
        if(nxt == p[x]) continue;
        tree[x].push_back(nxt);
        p[nxt] = x;
        dfs1(nxt);
        cntt++;
    }
    ans1 += (cntt / 2);
    if(x == 1 && (cntt % 2) == 1) ans1++;
}

bool dfs2(int x){
    l[x].clear();
    for(auto &nxt: tree[x]){
        if(!dfs2(nxt)) return false;
        //printf("bye %d -> %d: %d\n", nxt, x, bye[nxt]);
        l[x].push_back(bye[nxt]+1);
    }
    int m = (int)l[x].size();
    if(x == 1){
        sort(l[x].begin(), l[x].end());
        for(auto &val: l[x]){
            if(val > lim) return false;
        }
        if(m%2 == 0){
            for(int i=0;i<m;i++){
                if(l[x][i] + l[x][m-1-i] > lim) return false;
            }
        }else{
            for(int i=0;i<m-1;i++){
                if(l[x][i] + l[x][m-2-i] > lim) return false;
            }
        }
        return true;
    }
    if(m%2 == 0){
        l[x].push_back(0);
        m++;
    }
    sort(l[x].begin(), l[x].end());
    for(auto &val: l[x]){
        if(val > lim) return false;
    }
    bye[x] = 0;
    int ss = 0, ee = m-1, ans = m;
    while(ss <= ee){
        int mid = (ss + ee) / 2;
        int s = 0, e = m-1;
        bool can = true;
        while(s < e){
            if(s == mid) s++;
            if(e == mid) e--;
            if(s >= e) break;
            if(l[x][s] + l[x][e] > lim){
                can = false;
                break;
            }else{
                s++; e--;
            }
        }
        if(can){
            ans = mid;
            ee = mid - 1;
        }else{
            ss = mid + 1;
        }
    }
    if(ans == m) return false;
    bye[x] = l[x][ans];
    return true;
}

int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<n;i++){
        int aa, bb;
        scanf("%d%d", &aa, &bb);
        g[aa].push_back(bb);
        g[bb].push_back(aa);
    }
    p[1] = 1;
    dfs1(1);
    int ss = 0, ee = n+1, ans2 = n+1;
    while(ss <= ee){
        lim = (ss + ee) / 2;
        //printf("%d\n", lim);
        if(dfs2(1)){
            ans2 = lim;
            ee = lim - 1;
        }else{
            ss = lim + 1;
        }
    }
    printf("%d %d\n", ans1, ans2);
}
