Problem: 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

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
#define MAX 100002
int n;
vector<int> v[MAX];
int dp[MAX],cnt[MAX];
multiset<int> vv[MAX];
multiset<int> tmp[MAX];

bool can;
int mx;
vector<int> unf[MAX];
int C;
inline void dfs(int b, int pr = -1) {
	vv[b].clear();
	unf[b].clear();
	int deg = 0;
	for (int go : v[b]) {
		if (go == pr)continue;
		dfs(go, b);
		deg++;
		vv[b].insert(dp[go] + 1);
	}
	tmp[b] = vv[b];
	C += deg/2;
	if (deg % 2 && b == 0) {
		C++;
	}
	if (vv[b].size()) {
		int la = (*vv[b].rbegin());
		if (la > mx) {
			can = true;
			dp[b] = 0;
			return;
		}
	}
	while (vv[b].size() > 1) {
		int bc = *vv[b].rbegin();
		vv[b].erase(prev(vv[b].end()));
		int able = mx - bc;
		auto it = vv[b].upper_bound(able);
		if (it == vv[b].begin()) {
			unf[b].push_back(bc);
			continue;
		}
		it = prev(it);
		vv[b].erase(it);
	}
	if (vv[b].size()) {
		unf[b].push_back(*vv[b].begin());
	}
	vv[b].clear();
	if (deg%2&&unf[b].size() > 1) {
		can = true;
		dp[b] = 0;
		return;
	}
	if (deg % 2) {
		dp[b] = unf[b][0];
		return;
	}
	else {
		if (unf[b].size() == 0) {
			dp[b] = 0;
			return;
		}
		else {
			if (b == 0) {
				can = true;
				dp[b] = 0;
				return;
			}
			vv[b] = tmp[b];
			unf[b].clear();
			vv[b].erase(prev(vv[b].end()));
			while (vv[b].size() > 1) {
				int bc = *vv[b].rbegin();
				vv[b].erase(prev(vv[b].end()));
				int able = mx - bc;
				auto it = vv[b].upper_bound(able);
				if (it == vv[b].begin()) {
					unf[b].push_back(bc);
					continue;
				}
				it = prev(it);
				vv[b].erase(it);
			}
			if (vv[b].size()) {
				unf[b].push_back(*vv[b].begin());
			}
			if (unf[b].size() == 1) {
				dp[b] = unf[b][0];
				return;
			}
			can = true;
			dp[b] = 0;
			return;
		}
	}
}
bool ok(int len) {
	can = false;
	mx = len;
	C = 0;
	dfs(0);
	return can == false;
}
int main() {
	cin >> n;
	for (int i = 1; i < n; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	int mint = 1;
	int maxt = n;
	while (mint + 1 < maxt) {
		int mid = (mint + maxt) >> 1;
		if (ok(mid)) {
			maxt = mid;
		}
		else {
			mint = mid + 1;
		}
	}
	if (ok(mint)) {
		ok(mint);
		cout <<C<<" "<< mint << endl;
	}
	else {
		ok(maxt);
		cout << C<<" "<<maxt << endl;
	}
	return 0;
}