You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:

  * For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0. 



Note that you can put very large positive integers (like 10^{(10^{10})}).

It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.

<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).

<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0. 

What are the minimum and the maximum possible values of f for the given tree? Find and print both.

Input

The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.

The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.

Output

Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.

Examples

Input


6
1 3
2 3
3 4
4 5
5 6


Output


1 4


Input


6
1 3
2 3
3 4
4 5
4 6


Output


3 3


Input


7
1 2
2 7
3 4
4 7
5 6
6 7


Output


1 6

Note

In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3. 

<image>

In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>
#include <bits/stdc++.h>
using namespace std;
int deg[100005];
int p[100005];
vector<int> AdjList[100005];
int dist[100005];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    AdjList[a].push_back(b);
    AdjList[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  int root = 1;
  while (deg[root] == 1) {
    root++;
  }
  memset(dist, -1, sizeof(dist));
  memset(p, -1, sizeof(p));
  queue<int> q;
  q.push(root);
  dist[root] = 0;
  p[root] = -1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : AdjList[u]) {
      if (dist[v] == -1) {
        dist[v] = dist[u] + 1;
        p[v] = u;
        q.push(v);
      }
    }
  }
  int num_even = 0;
  int num_odd = 0;
  set<int> s;
  int max_edges = N - 1;
  for (int i = 1; i <= N; i++) {
    if (deg[i] == 1) {
      if (s.count(p[i])) {
        max_edges--;
      } else {
        s.insert(p[i]);
      }
      if (dist[i] % 2 == 0) {
        num_even++;
      } else {
        num_odd++;
      }
    }
  }
  int ans1 = 3;
  if (num_even == 0 || num_odd == 0) {
    ans1 = 1;
  }
  printf("%d %d", ans1, max_edges);
  return 0;
}
