#include <bits/stdc++.h>
using namespace std;
long long int sum = 0;
std::vector<std::vector<long long int>> adj(2000000);
long long int cnt[2], vis[2000000];
void dfs(long long int v, long long int c) {
  vis[v] = 1;
  if (adj[v].size() == 1) {
    if (c % 2 == 0)
      cnt[0]++;
    else {
      cnt[1]++;
    }
    return;
  }
  long long int p = 0, i;
  for (i = 0; i < adj[v].size(); i++) {
    if (adj[adj[v][i]].size() == 1) {
      p++;
    }
  }
  for (i = 0; i < adj[v].size(); i++) {
    if (!vis[adj[v][i]]) dfs(adj[v][i], c + 1);
  }
  if (p > 0) sum = sum - (p - 1);
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int n, i, in = -1;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    if (adj[x].size() > 1) in = x;
    if (adj[y].size() > 1) in = y;
  }
  sum = n - 1;
  dfs(in, 0);
  long long int mini = 1;
  if (cnt[0] > 0 && cnt[1] > 0) {
    mini = 3;
  }
  cout << mini << " " << sum << endl;
}

### Prompt

Please formulate a cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int sum = 0;
std::vector<std::vector<long long int>> adj(2000000);
long long int cnt[2], vis[2000000];
void dfs(long long int v, long long int c) {
  vis[v] = 1;
  if (adj[v].size() == 1) {
    if (c % 2 == 0)
      cnt[0]++;
    else {
      cnt[1]++;
    }
    return;
  }
  long long int p = 0, i;
  for (i = 0; i < adj[v].size(); i++) {
    if (adj[adj[v][i]].size() == 1) {
      p++;
    }
  }
  for (i = 0; i < adj[v].size(); i++) {
    if (!vis[adj[v][i]]) dfs(adj[v][i], c + 1);
  }
  if (p > 0) sum = sum - (p - 1);
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int n, i, in = -1;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    if (adj[x].size() > 1) in = x;
    if (adj[y].size() > 1) in = y;
  }
  sum = n - 1;
  dfs(in, 0);
  long long int mini = 1;
  if (cnt[0] > 0 && cnt[1] > 0) {
    mini = 3;
  }
  cout << mini << " " << sum << endl;
}

```