#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> adj[n + 1];
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<bool> lf(n + 1);
  long long lfc = 0, start = INT_MAX;
  for (long long i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      start = min(start, i);
      lfc++;
      lf[i] = 1;
    }
  }
  long long mn = 1, mx = n - 1;
  vector<long long> d(n + 1, -1);
  queue<long long> q;
  q.push(start);
  d[start] = 0;
  while (!q.empty()) {
    long long u = q.front();
    q.pop();
    long long cnt = 0;
    for (long long i = 0; i < adj[u].size(); i++) {
      long long v = adj[u][i];
      if (lf[v]) cnt++;
      if (d[v] == -1) {
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
    if (cnt >= 2) mx -= (cnt - 1);
  }
  for (long long i = 1; i <= n; i++)
    if (lf[i] and d[i] % 2 == 1 and d[i] > 1) mn = 3;
  cout << mn << ' ' << mx << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int qu = 1;
  for (int cas = 1; cas <= qu; cas++) {
    solve();
  }
}

### 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;
void solve() {
  long long n;
  cin >> n;
  vector<long long> adj[n + 1];
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<bool> lf(n + 1);
  long long lfc = 0, start = INT_MAX;
  for (long long i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      start = min(start, i);
      lfc++;
      lf[i] = 1;
    }
  }
  long long mn = 1, mx = n - 1;
  vector<long long> d(n + 1, -1);
  queue<long long> q;
  q.push(start);
  d[start] = 0;
  while (!q.empty()) {
    long long u = q.front();
    q.pop();
    long long cnt = 0;
    for (long long i = 0; i < adj[u].size(); i++) {
      long long v = adj[u][i];
      if (lf[v]) cnt++;
      if (d[v] == -1) {
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
    if (cnt >= 2) mx -= (cnt - 1);
  }
  for (long long i = 1; i <= n; i++)
    if (lf[i] and d[i] % 2 == 1 and d[i] > 1) mn = 3;
  cout << mn << ' ' << mx << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int qu = 1;
  for (int cas = 1; cas <= qu; cas++) {
    solve();
  }
}

```