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;
#define ran 100010
int n,res,W;
vector<int> e[ran];
int sz[ran];
void dfs1(int x,int fat){
	int cnt = 0;
	sz[x] = 0;
	for(vector<int>::iterator it = e[x].begin(); it != e[x].end(); it++){
		int y = *it;
		if(y == fat)continue;
		dfs1(y, x);
		cnt ++;
		sz[x] += sz[y];
	}
	sz[x] += cnt / 2;
	if(x == fat)sz[x] += cnt % 2;
}
bool dfs(int x,int fat){
	vector<int> e;
	for(vector<int>::iterator it = ::e[x].begin(); it != ::e[x].end(); it++){
		int y = *it;
		if(y == fat)continue;
		if(!dfs(y, x))return false;
		e.push_back(sz[y] + 1);
	}
	if(x == fat){
		if(e.size() % 2)e.push_back(0);
		sort(e.begin(), e.end());
		int L = e.size();
		for(int i=0,j=L-1; i<j; i++,j--)
			if(e[i] + e[j] > W)
				return false;
		return true;
	}
	if(e.size() % 2 == 0)e.push_back(0);
	sort(e.begin(), e.end());
	int L = e.size();
	int lo = 0, hi = L;
	while(lo != hi){
		int mi = (lo + hi)/2;
		bool flag = true;
		for(int i=0,j=L-1;;){
			if(i==mi)i++;
			if(j==mi)j--;
			if(i>j)break;
			if(e[i] + e[j] > W){
				flag = false;
				break;
			}
			i++;j--;
		}
		if(flag)
			hi = mi;
		else
			lo = mi+1;
	}
	if(hi == L)return false;
	sz[x] = e[lo];
	return true;
}
int main(){
	scanf("%d",&n);
	for(int i=1; i<n; i++){
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs1(1,1);
	printf("%d ",sz[1]);
	int lo = 1, hi = n-1;
	while(lo != hi){
		W = (lo + hi)/2;
		if(dfs(1, 1))
			hi = W;
		else
			lo = W+1;
	}
	printf("%d\n",lo);
	return 0;
}
