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;
bool hasEvenLeaf[100001];
bool hasOddLeaf[100001];
bool seen[100001];
bool isLeaf[100001];
int leafCount[100001];
vector<long long> adj[100001];
void bfs(int x) {
  if (adj[x].size() == 1) {
    isLeaf[x] = true;
    hasEvenLeaf[x] = true;
  }
  seen[x] = true;
  for (auto it : adj[x]) {
    if (!seen[it]) {
      bfs(it);
    }
  }
  for (auto it : adj[x]) {
    if (hasEvenLeaf[it]) hasOddLeaf[x] = true;
    if (hasOddLeaf[it]) hasEvenLeaf[x] = true;
    if (hasEvenLeaf[x]) hasOddLeaf[it] = true;
    if (hasOddLeaf[x]) hasEvenLeaf[it] = true;
  }
}
void bfs2(int x) {
  if (isLeaf[x]) {
    hasEvenLeaf[x] = true;
  }
  seen[x] = true;
  for (auto it : adj[x]) {
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long a, b;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  memset(hasOddLeaf, false, sizeof(hasOddLeaf));
  memset(hasEvenLeaf, false, sizeof(hasEvenLeaf));
  memset(seen, false, sizeof(seen));
  memset(isLeaf, false, sizeof(isLeaf));
  memset(leafCount, 0, sizeof(leafCount));
  bfs(0);
  long long minAns = 1;
  for (int i = 0; i < n; i++) {
    if (hasEvenLeaf[i] && hasOddLeaf[i]) {
      minAns = 3;
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    for (auto it : adj[i]) {
      if (isLeaf[it]) leafCount[i]++;
    }
  }
  long long maxAns = n - 1;
  for (int i = 0; i < n; i++) {
    maxAns -= max((leafCount[i] - 1), 0);
  }
  cout << minAns << " " << maxAns << '\n';
}
