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;
typedef long long ll;
const int N=1e5+5;
int n,tot,ans,dp[N],sum[N],head[N],nex[N<<1],to[N<<1];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
void dfs(int u,int p)
{
    int tot=0;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        dfs(v,u);
        dp[u]+=dp[v];
        tot++;
    }
    dp[u]+=tot/2;
    if(u==1) dp[u]+=tot%2;
}
int top,s[N];
bool dfs(int u,int p,int m)
{
    sum[u]=0;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        if(!dfs(v,u,m)) return false;
    }
    top=0;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        s[++top]=sum[v]+1;
    }
    if(u==1)
    {
        if(top&1) s[++top]=0;
        sort(s+1,s+1+top);
        for(int i=1,j=top;i<j;i++,j--)
            if(s[i]+s[j]>m) return false;
        return true;
    }
    if(top%2==0) s[++top]=0;
    sort(s+1,s+1+top);
    int l=1,r=top,ans=-1;
    while(l<=r)
    {
        int mid=l+r>>1;
        bool flag=true;
        for(int i=1,j=top;i<j;i++,j--)
        {
            if(i==mid) i++;
            if(j==mid) j--;
            if(i>j) break;
            if(s[i]+s[j]>m) {flag=false;break;}
        }
        if(flag) ans=s[mid],r=mid-1;
        else l=mid+1;
    }
    if(ans==-1) return false;
    sum[u]=ans;
    return true;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    dfs(1,0);
    int l=1,r=n,ans;
    while(l<=r)
    {
        int m=l+r>>1;
        if(dfs(1,0,m)) ans=m,r=m-1;
        else l=m+1;
    }
    printf("%d %d\n",dp[1],ans);
}
