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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 100005;

class Edge {
public:
  int nxt, to;
} e[MAX_N << 1];

int head[MAX_N], cnt;
void addedge(int u, int v) {
  e[++cnt] = (Edge){head[u], v}, head[u] = cnt;
  e[++cnt] = (Edge){head[v], u}, head[v] = cnt;
}

int N, length[MAX_N];
int que[MAX_N], top, ans2, mid, ok, ans1;

bool check(int pos) {
  for (int i = 1, j = top; i < j; ++i, --j) {
    if (i == pos) i++;
    if (j == pos) j--;
    if (i < j && que[i] + que[j] > mid) return false;
  }
  return true;
}

void dfs(int u, int v) {
  if (!ok) return;
  
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].to != v)
      dfs(e[i].to, u);
  if (!ok) return;
  
  top = 0;
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].to != v)
      que[++top] = length[e[i].to];
  sort(que + 1, que + top + 1);
  
  if (top % 2 == 0) {
    bool flag = 1;
    for (int i = 1; i <= top / 2; ++i)
      if (que[i] + que[top - i + 1] > mid) {
	flag = 0;
	break;
      }
    if (flag) {
      length[u] = 1, ans1 += top / 2;
      return;
    }
    if (v == 0) {
      ok = 0;
      return;
    }
    top--, ans1++;
  }
  que[top + 1] = 0;
  if (top & 1) {
    int l = 1, r = top, res = -1;
    while (l <= r) {
      int tmid = (l + r) >> 1;
      if (check(tmid)) res = tmid, r = tmid - 1;
      else l = tmid + 1;
    }
    if (res == -1) ok = 0;
    if (que[res] == mid && v != 0) ok = 0;
    length[u] = que[res] + 1, ans1 += top / 2;
    if (v == 0) ans1++;
  }
}

int main() {
  scanf("%d", &N);
  for (int i = 1, u, v; i < N; ++i) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
  }
  int l = 1, r = N, ans2 = N;
  while (l <= r) {
    ans1 = 0;
    mid = (l + r) >> 1;
    ok = 1, dfs(1, 0);
    if (ok) ans2 = mid, r = mid - 1;
    else l = mid + 1;
  }
  
  ans1 = 0, ok = 1, mid = ans2;
  dfs(1, 0);
  printf("%d %d\n", ans1, ans2);
  return 0;
}