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>

#define pb push_back
#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

const int N = 1e5 + 10;

int n, deg[N];
vector<int> G[N];

bool check(const vector<int>& A, int x, int m){
	int l = 0, r = int(A.size()) - 1;
	while(l < r){
		if(l == x) ++l;
		if(r == x) --r;
		if(A[l] + A[r] > m) return false;
		++l, --r;
	}
	return true;
}

int dp[N];

bool DFS(int o, int f, int lim){

	vector<int> val;
	
	int ch = 0;
	for(int v : G[o]){
		if(v == f) continue;
		++ch;
		if(!DFS(v, o, lim)) return false;
		val.pb(dp[v]);
	}

	if(ch % 2 == 0) val.pb(0);
	sort(val.begin(), val.end());

	int L = 0, R = int(val.size()) - 1;
	if(!check(val, R, lim)) return false;

	while(L < R){
		int mid = (L + R) >> 1;
		if(check(val, mid, lim)) R = mid;
		else L = mid + 1;
	}
	dp[o] = val[L] + 1;

	return true;
}

int main(){
	
	scanf("%d", &n);
	For(i, 2, n){
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v), G[v].pb(u);
		deg[u]++, deg[v]++;
	}

	int root = 0, a = 0;
	For(i, 1, n) if(deg[i] == 1){
		root = i;
		break;
	}
	For(i, 1, n) if(deg[i] & 1) ++a;

	int L = 1, R = n - 1;
	while(L < R){
		int mid = (L + R) >> 1;
		if(DFS(root, 0, mid) && dp[root] <= mid + 1) R = mid;
		else L = mid + 1;
	}

	printf("%d %d\n", a / 2, L);

	return 0;
}