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<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

//#ifdef LOCAL
//#define eprintf(...) fprintf(stderr, __VA_ARGS__)
//#else
//#define NDEBUG
//#define eprintf(...) do {} while (0)
//#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

int N;
VI G[100011];
VI ord;
int par[100011];
int dp[100011];
VI t, st;

bool ok(int mid) {
    REP (i_, N) {
	int v = ord[N-1-i_];
	t.clear();
	EACH (e, G[v]) if (*e != par[v]) {
	    if (dp[*e]+1 > mid) return false;
	    t.push_back(dp[*e]+1);
	}

	if (v == 0) {
	    int m = t.size()/2;
	    sort(t.begin(), t.end());
	    REP (j, m) if (t[j] + t[m*2-1-j] > mid) return false;
	    return true;
	} else {
	    if (t.size() % 2 == 0) t.push_back(0);
	    sort(t.begin(), t.end());
	    st.clear();
	    int pos = 0;
	    dp[v] = -1;
	    for (int j=t.size()-1; j>=0; j--) {
		int c = t[j];
		if (c == -1) break;
		while (pos < j && t[pos] + c <= mid) {
		    st.push_back(t[pos]);
		    t[pos] = -1;
		    pos++;
		}
		if (st.empty()) {
		    if (dp[v] == -1) dp[v] = c;
		    else return false;
		} else {
		    st.pop_back();
		}
	    }

	    if (dp[v] == -1) {
		dp[v] = st[0];
	    }
	}
    }

    assert(false);
}

void MAIN() {
    scanf("%d", &N);
    REP (i, N-1) {
	int x, y;
	scanf("%d%d", &x, &y);
	x--; y--;
	G[x].push_back(y);
	G[y].push_back(x);
    }

    ord.reserve(N);
    ord.push_back(0);
    REP (i, N) {
	int v = ord[i];
	EACH (e, G[v]) if (*e != par[v]) {
	    par[*e] = v;
	    ord.push_back(*e);
	}
    }

    int A = 1;
    REP (i, N) {
	A += ((int)G[i].size() - 1) / 2;
    }

    int lo = 0, hi = N-1;
    while (hi - lo > 1) {
	int mid = (lo + hi) / 2;
	(ok(mid)? hi: lo) = mid;
    }
    printf("%d %d\n", A, hi);
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}