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> lis[100100];
vector<int> cdp[100100];
int dp[100100];

bool dfs(int here, int p, int t) {
    cdp[here].clear(); int i, j;
    for (auto &there : lis[here]) {
        if (there==p) continue;
        if (!dfs(there,here,t)) return false;
        cdp[here].push_back(dp[there]+1);
    }
    sort(cdp[here].begin(),cdp[here].end());
    if (cdp[here].empty()) {
        dp[here] = 0;
        return true;
    }
    if (p<0) {
        for (i=0,j=(int)cdp[here].size()/2*2-1;i<j;i++,j--) {
            if (cdp[here][i]+cdp[here][j]>t) return false;
        }
        if (cdp[here].back()>t) return false;
        return true;
    }
    if (cdp[here].size()%2==0) {
        for (i=0,j=(int)cdp[here].size()-1;i<j;i++,j--) {
            if (cdp[here][i]+cdp[here][j]>t) break;
        }
        if (i>=j) {
            dp[here] = 0;
            return true;
        }
        if (cdp[here].back()>t) return false;
    }
    int cur = 0;
    for (i=1,j=((int)cdp[here].size()-1)/2*2;i<j;i++,j--) {
        if (cdp[here][i]+cdp[here][j]>t) {
            cur = i;
            if (cdp[here][i-1]+cdp[here][j]>t) break;
        }
    }
    if (i>=j) {
        dp[here] = cdp[here][cur];
        return true;
    }
    cur = ((int)cdp[here].size()-1)/2*2; int fl = 0;
    for (i=0,j=((int)cdp[here].size()-1)/2*2-1;i<j;i++,j--) {
        if (cdp[here][i]+cdp[here][j]>t) return false;
        if (cdp[here][i]+cdp[here][j+1]<=t&&!fl) cur = j;
        else fl = 1;
    }
    dp[here] = cdp[here][cur];
    return true;
}

bool ok(int t) {
    return dfs(0,-1,t);
}

int main() {
    int i;

    scanf("%d",&n);
    for (i=0;i<n-1;i++){
        int a, b;
        scanf("%d%d",&a,&b);--a;--b;
        lis[a].push_back(b);
        lis[b].push_back(a);
    }
    int r1 = 1;
    for (i=0;i<n;i++) r1 += ((int)lis[i].size()-1)/2;
    printf("%d ",r1);
    int s = 0, e = n;
    while(s<=e) {
        int m = (s+e)>>1;
        if (ok(m)) e = m-1;
        else s = m+1;
    }
    printf("%d\n",s);

    return 0;
}
