#include <bits/stdc++.h>
using namespace std;
const int dx[8] = {-1, 0, 0, 1, 1, 1, -1, -1};
const int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
const int N = 1e5 + 5;
vector<int> adj[N + 1];
int maxf = 0, minf = 1;
int visit[N + 1], parent[N + 1], parentofleaf[N + 1];
void dfs(int v, int count) {
  visit[v] = 1;
  if (parentofleaf[v] == 1) {
    if (count % 2 == 1) minf = 3;
  }
  int no = 0;
  for (auto u : adj[v]) {
    if (adj[u].size() == 1) no++;
    if (visit[u] == -1 && adj[u].size() > 1) {
      dfs(u, count + 1);
    }
  }
  if (no > 0) maxf -= (no - 1);
}
void initiall(int n) {
  for (int i = 0; i <= n; i++) {
    adj[i].clear();
    visit[i] = -1;
    parent[i] = 0;
    parentofleaf[i] = 0;
  }
}
void solve() {
  int n, i;
  cin >> n;
  int a, b;
  initiall(n);
  for (i = 1; i < n; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    parent[a] = b;
    parent[b] = a;
  }
  for (i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      parentofleaf[parent[i]] = 1;
    }
  }
  maxf = n - 1;
  for (i = 1; i <= n; i++) {
    if (visit[i] == -1 && parentofleaf[i] == 1) dfs(i, 0);
  }
  cout << minf << " " << maxf << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  for (int j = 1; j <= tc; j++) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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 dx[8] = {-1, 0, 0, 1, 1, 1, -1, -1};
const int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
const int N = 1e5 + 5;
vector<int> adj[N + 1];
int maxf = 0, minf = 1;
int visit[N + 1], parent[N + 1], parentofleaf[N + 1];
void dfs(int v, int count) {
  visit[v] = 1;
  if (parentofleaf[v] == 1) {
    if (count % 2 == 1) minf = 3;
  }
  int no = 0;
  for (auto u : adj[v]) {
    if (adj[u].size() == 1) no++;
    if (visit[u] == -1 && adj[u].size() > 1) {
      dfs(u, count + 1);
    }
  }
  if (no > 0) maxf -= (no - 1);
}
void initiall(int n) {
  for (int i = 0; i <= n; i++) {
    adj[i].clear();
    visit[i] = -1;
    parent[i] = 0;
    parentofleaf[i] = 0;
  }
}
void solve() {
  int n, i;
  cin >> n;
  int a, b;
  initiall(n);
  for (i = 1; i < n; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    parent[a] = b;
    parent[b] = a;
  }
  for (i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      parentofleaf[parent[i]] = 1;
    }
  }
  maxf = n - 1;
  for (i = 1; i <= n; i++) {
    if (visit[i] == -1 && parentofleaf[i] == 1) dfs(i, 0);
  }
  cout << minf << " " << maxf << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  for (int j = 1; j <= tc; j++) {
    solve();
  }
  return 0;
}

```