#include <bits/stdc++.h>
using namespace std;
struct go {
  long long a, b;
};
const long long N = 1e5, inf = 2e9;
vector<long long> g[N];
pair<long long, long long> ch[N];
long long mn, mx;
void dfs(long long v, long long p) {
  long long cnt = 0;
  if (g[v].size() == 1) ch[v].second = 1;
  long long nw = 0, GG = 0;
  for (long long to : g[v]) {
    if (to != p) {
      dfs(to, v);
      if (g[to].size() == 1) ++cnt;
      GG++;
      if (ch[v].first && ch[to].first) nw = 1;
      if (ch[v].second && ch[to].second) nw = 1;
      if (ch[to].first) {
        ch[v].second = 1;
      }
      if (ch[to].second) {
        ch[v].first = 1;
      }
    }
  }
  if (nw) mn = 3;
  mx -= max((cnt - 1), (long long)0);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; ++i) ch[i] = {0, 0};
  long long n;
  cin >> n;
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    g[--u].push_back(--v);
    g[v].push_back(u);
  }
  long long st = 0;
  mn = 1, mx = n - 1;
  for (long long i = 0; i < n; ++i) {
    if (g[i].size() >= 2) {
      st = i;
    }
  }
  dfs(st, -1);
  cout << mn << " " << mx;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
struct go {
  long long a, b;
};
const long long N = 1e5, inf = 2e9;
vector<long long> g[N];
pair<long long, long long> ch[N];
long long mn, mx;
void dfs(long long v, long long p) {
  long long cnt = 0;
  if (g[v].size() == 1) ch[v].second = 1;
  long long nw = 0, GG = 0;
  for (long long to : g[v]) {
    if (to != p) {
      dfs(to, v);
      if (g[to].size() == 1) ++cnt;
      GG++;
      if (ch[v].first && ch[to].first) nw = 1;
      if (ch[v].second && ch[to].second) nw = 1;
      if (ch[to].first) {
        ch[v].second = 1;
      }
      if (ch[to].second) {
        ch[v].first = 1;
      }
    }
  }
  if (nw) mn = 3;
  mx -= max((cnt - 1), (long long)0);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; ++i) ch[i] = {0, 0};
  long long n;
  cin >> n;
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    g[--u].push_back(--v);
    g[v].push_back(u);
  }
  long long st = 0;
  mn = 1, mx = n - 1;
  for (long long i = 0; i < n; ++i) {
    if (g[i].size() >= 2) {
      st = i;
    }
  }
  dfs(st, -1);
  cout << mn << " " << mx;
}

```