#include <bits/stdc++.h>
using namespace std;
const int N = 1000 * 100 + 3;
int n, ans1, h[N];
bool par[N], mark[N], vis[N];
vector<int> adj[N];
void input() {
  cin >> n;
  for (int i = 0; i + 1 < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[--u].push_back(--v);
    adj[v].push_back(u);
  }
}
void dfs2(int u) {
  vis[u] = true;
  for (auto v : adj[u])
    if (vis[v] == false) {
      h[v] = h[u] + 1;
      dfs2(v);
    }
  if (adj[u].size() == 1 && h[u] % 2 == 0) ans1 = 3;
}
void dfs1(int u) {
  mark[u] = true;
  for (auto v : adj[u])
    if (mark[v] == false) {
      if (adj[v].size() == 1) par[u] = true;
      dfs1(v);
    }
}
void solve() {
  ans1 = 1;
  for (int i = 0; i < n; i++)
    if (adj[i].size() == 1) {
      dfs1(adj[i][0]);
      dfs2(adj[i][0]);
      return;
    }
}
void output() {
  cout << ans1 << ' ';
  int ans = n - 1;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) ans--;
    if (par[i] == true) ans++;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(), solve(), output();
  return 0;
}

### 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;
const int N = 1000 * 100 + 3;
int n, ans1, h[N];
bool par[N], mark[N], vis[N];
vector<int> adj[N];
void input() {
  cin >> n;
  for (int i = 0; i + 1 < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[--u].push_back(--v);
    adj[v].push_back(u);
  }
}
void dfs2(int u) {
  vis[u] = true;
  for (auto v : adj[u])
    if (vis[v] == false) {
      h[v] = h[u] + 1;
      dfs2(v);
    }
  if (adj[u].size() == 1 && h[u] % 2 == 0) ans1 = 3;
}
void dfs1(int u) {
  mark[u] = true;
  for (auto v : adj[u])
    if (mark[v] == false) {
      if (adj[v].size() == 1) par[u] = true;
      dfs1(v);
    }
}
void solve() {
  ans1 = 1;
  for (int i = 0; i < n; i++)
    if (adj[i].size() == 1) {
      dfs1(adj[i][0]);
      dfs2(adj[i][0]);
      return;
    }
}
void output() {
  cout << ans1 << ' ';
  int ans = n - 1;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) ans--;
    if (par[i] == true) ans++;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(), solve(), output();
  return 0;
}

```