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 <cstdio>
#include <algorithm>
#define FOR(i, l, r) for(int i = l; i <= r; ++i)

using namespace std;

const int N = 200010;

struct edge{int to, next;} e[N];
int du[N], head[N], a[N], f[N];
int n, x, y, cnt, tot, ans1, ans2, rt;
bool fl;

void ins(int x, int y)
{
    e[++cnt].to = y; e[cnt].next = head[x]; head[x] = cnt; ++du[x];
}

bool check2(int mid, int len)
{
    int i = 1, j = tot;
    if (i == mid) ++i;
    if (j == mid) --j;

    for(; i < j;)
    {
        if (a[i] + a[j] > len) return 0;
        ++i, --j;
        if (i == mid) ++i;
        if (j == mid) --j;
    }
    return 1;
}

void dfs(int x, int y, int len)
{
    for(int i = head[x]; i; i = e[i].next)
        if (e[i].to != y) dfs(e[i].to, x, len);
    tot = 0;
    for(int i = head[x]; i; i = e[i].next)
        if (e[i].to != y) a[++tot] = f[e[i].to];
    if (!tot) {f[x] = 1; return;}

    if (!(tot & 1)) a[++tot] = 0;
    sort(a + 1, a + tot + 1);

    int l = 1, r = tot, U = n + 1;
    while (l <= r)
    {
        int mid = (l + r) >> 1;
        if (check2(mid, len)) U = a[mid], r = mid - 1; else l = mid + 1;
    }
    if (U == n + 1) fl = 1;
    f[x] = U + 1;
}

bool check(int len)
{
    bool fl = 0; dfs(rt, 0, len);
    return !fl && f[rt] <= len + 1;
}

int main()
{
    scanf("%d", &n);
    FOR(i, 2, n)
    {
        scanf("%d%d", &x, &y);
        ins(x, y); ins(y, x);
    }
    FOR(i, 1, n) if (du[i] & 1) rt = i, ++ans1;
    ans1 >>= 1;
    int l = 1, r = n;
    while (l <= r)
    {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid - 1, ans2 = mid; else l = mid + 1;
    }
    printf("%d %d\n", ans1, ans2);
    return 0;
}
