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;
const int N = 1e5 + 10;
int nt[N << 1], to[N << 1], head[N], cnt = 0;

void add(int x, int y) {
  to[++cnt] = y;
  nt[cnt] = head[x];
  head[x] = cnt;
}

int f[N], g[N], ans = 0;
void predfs(int u, int fa) {
  int s = 0, pos = 0, tmp = 0, mx = 0;
  for(int i = head[u]; i; i = nt[i]) {
    if(to[i] == fa) continue;
    predfs(to[i], u);
    ans = max(ans, mx + g[to[i]] + 1);
    mx = max(mx, g[to[i]] + 1);
    pos += f[to[i]];
    tmp++;
  }
  g[u] = mx;
  f[u] = pos + (tmp / 2);
  if(u == 1) f[u] += tmp & 1;
  if(!tmp) f[u] = 0;
}

int lmt, flg, up[N], A, B, num, st[N];
multiset<int> S;
multiset<int> :: iterator sit;
void dfs(int u, int fa) {
  if(!flg) return ;
  for(int i = head[u]; i; i = nt[i]) {
    if(to[i] == fa) continue;
    dfs(to[i], u);
  }
//  printf("%d %d\n", u, fa);
  for(int i = head[u]; i; i = nt[i]) {
    if(to[i] == fa) continue;
    if(f[to[i]] + 1 > lmt) {
      flg = 0;
      return ;
    }
//    printf("%d %d num = %d\n", to[i], f[to[i]] + 1, num);
    S.insert(f[to[i]] + 1);
  }
  int top = 0;
  while(S.size() > 1) {
    int pos = *S.rbegin();
    S.erase(S.find(pos));
    if(*S.rbegin() + pos <= lmt) {
      num++;
      int tmp = *S.rbegin();
      S.erase(S.find(tmp));
    } else if(*S.begin() + pos > lmt) {
      st[++top] = pos;
    } else {
      sit = S.upper_bound(lmt - pos);
      sit--;
      num++;
      S.erase(sit);
    }
  }
  if(S.size()) {
    st[++top] = *S.begin();
    S.clear();
  }
  f[u] = 0;
  if(top) {
    sort(st + 1, st + 1 + top);
    f[u] = st[1];
    num += top - 1;
  }
//  if(u == 2) printf("qswl%d\n", f[u]);
  if(!fa && f[u]) num++;
}

bool check(int x) {
  flg = 1;
  lmt = x, num = 0;
  dfs(1, 0);
//  printf("qwq%d %d %d\n", x, num, flg);
  return flg && num <= A;
}

int main() {
  int n;
  scanf("%d", &n);
  for(int i = 2, x, y; i <= n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
  }
  memset(f, 0x3f, sizeof f);
  predfs(1, 0);
  A = f[1];
  int l = 1, r = ans;
  while(l <= r) {
    int mid = (l + r) >> 1;
    if(check(mid)) r = mid - 1, B = mid;
    else l = mid + 1;
  }
//  check(2);
  printf("%d %d\n", A, B);
  return 0;
}